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# Word Problems on Integration

Prerequisite Topics:
Factoring
Partial Fractions
Exponents and Logarithms
Trigonometry
Differential Calculus

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Indicate the method(s) you used as applicable
Show all work

(1.) Evaluate $\displaystyle\int -7dx$

Power Rule

$\displaystyle\int -7dx \\[5ex] = \displaystyle\int -7 * 1 * dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \displaystyle\int -7x^0dx \\[5ex] = -7 \displaystyle\int x^0dx \\[5ex] = -7 * \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = -7 * \dfrac{x^1}{1} + C \\[5ex] = -7x + C$
(2.) Determine the antiderivative of $\cot x$

Integration by Algebraic Substitution

$\displaystyle\int \cot xdx \\[5ex] \cot x = \dfrac{\cos x}{\sin x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \sin x \\[3ex] \dfrac{dp}{dx} = \cos x \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{\cos x} \\[5ex] dx = \dfrac{dp}{\cos x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \cot xdx = \displaystyle\int \dfrac{\cos x}{\sin x} dx \\[5ex] = \displaystyle\int \dfrac{\cos x}{p} * \dfrac{dp}{\cos x} \\[5ex] = \displaystyle\int \dfrac{dp}{p} \\[5ex] = \ln p + C \\[3ex] = \ln \sin x + C$
(3.) JAMB Evaluate $\displaystyle\int_1^3 (x^2 - 1)dx$

$A.\:\: -6\dfrac{2}{3} \\[5ex] B.\:\: 6\dfrac{2}{3} \\[5ex] C.\:\: \dfrac{2}{3} \\[5ex] D.\:\: -\dfrac{2}{3} \\[5ex]$

Power Rule

$\displaystyle\int_1^3 (x^2 - 1)dx \\[5ex] \displaystyle (x^2 - 1)dx = \displaystyle\int x^2dx - \displaystyle\int 1dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \dfrac{x^{2 + 1}}{2 + 1} - \displaystyle\int x^0dx \\[5ex] = \dfrac{x^3}{3} - \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = \dfrac{x^3}{3} - x + C \\[5ex] \rightarrow \\[3ex] \displaystyle\int_1^3 (x^2 - 1)dx = \left[\dfrac{x^3}{3} - x\right]_1^3 \\[5ex] = \left(\dfrac{3^3}{3} - 3\right) - \left(\dfrac{1^3}{3} - 1\right) \\[5ex] = (9 - 3) - \left(\dfrac{1}{3} - \dfrac{3}{3}\right) \\[5ex] = 6 - \left(-\dfrac{2}{3}\right) \\[5ex] = 6 + \dfrac{2}{3} \\[5ex] = 6\dfrac{2}{3}$
(4.) JAMB Evaluate $\displaystyle\int \sin 3xdx$

$A.\:\: -\dfrac{1}{3}\cos 3x + c \\[5ex] B.\:\: \dfrac{1}{3}\cos 3x + c \\[5ex] C.\:\: \dfrac{2}{3}\cos 3x + c \\[5ex] C.\:\: -\dfrac{2}{3}\cos 3x + c \\[5ex]$

Integration by Algebraic Substitution

$\displaystyle\int \sin 3xdx \\[5ex] Let\:\: p = 3x \\[3ex] \dfrac{dp}{dx} = 3 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{3} \\[5ex] dx = \dfrac{dp}{3} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \sin 3xdx = \displaystyle\int \sin p * \dfrac{dp}{3} \\[5ex] = \dfrac{1}{3} * \displaystyle\int \sin pdp \\[5ex] = \dfrac{1}{3} * -\cos p + C \\[5ex] = \dfrac{1}{3} * - \cos 3x + C \\[3ex] = -\dfrac{1}{3}\cos 3x + c$
(5.) WASSCE:FM

(a) Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions

(b) Use the answer in (a) to evaluate $\displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3}dx$

$(a.) \\[3ex] \dfrac{x + 6}{(x + 1)^3} \\[5ex]$ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator

$\dfrac{x + 6}{(x + 1)^3} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} \\[5ex] = \dfrac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3} \\[5ex] = \dfrac{A[(x + 1)(x + 1)] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + x + x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + A + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + Bx + A + B + C}{(x + 1)^3} \\[5ex]$ Denominators are the same
Equate the numerators

$x + 6 = Ax^2 + 2Ax + Bx + A + B + C \\[3ex] Swap \\[3ex] Ax^2 + 2Ax + Bx + A + B + C = x + 6 \\[3ex] Ax^2 + x(2A + B) + A + B + C = 0x^2 + x + 6 \\[3ex] \implies \\[3ex] A = 0...eqn.(1) \\[3ex] 2A + B = 1...eqn.(2) \\[3ex] A + B + C = 6...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 2(0) + B = 1 \\[3ex] 0 + B = 1 \\[3ex] B = 1 - 0 \\[3ex] B = 1 \\[3ex] Substitute\;\;0\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 0 + 1 + C = 6 \\[3ex] 1 + C = 6 \\[3ex] C = 6 - 1 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{0}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \dfrac{x + 6}{(x + 1)^3} = 0 + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \therefore \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] = \dfrac{1(x + 1) + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 1 + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 6}{(x + 1)^3} \\[5ex] = LHS \\[3ex]$ Integration by Partial Fractions

$(b.) \\[3ex] \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx \\[5ex] = \displaystyle\int_1^2 \left[\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}\right] dx \\[5ex] = \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] Let\;\; p = x + 1 \\[3ex] \dfrac{dp}{dx} = 1 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{1} = 1 \\[5ex] dx = dp \\[3ex] \displaystyle\int \dfrac{1}{(x + 1)^2} dx \\[5ex] = \displaystyle\int \dfrac{1}{p^2} dp \\[5ex] = \displaystyle\int p^{-2} dp \\[5ex] = \dfrac{p^{-2 + 1}}{-2 + 1} \\[5ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] = -\dfrac{1}{x + 1} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx \\[5ex] = \left[-\dfrac{1}{x + 1}\right]_1^2 \\[5ex] = -\dfrac{1}{2 + 1} - -\dfrac{1}{1 + 1} \\[5ex] = -\dfrac{1}{3} + \dfrac{1}{2} \\[5ex] = \dfrac{-2 + 3}{6} \\[5ex] = \dfrac{1}{6} \\[7ex] \displaystyle\int \dfrac{5}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{p^3} dp \\[5ex] = 5\displaystyle\int p^{-3} dp \\[5ex] = 5 * \dfrac{p^{-3 + 1}}{-3 + 1} \\[5ex] = 5 * \dfrac{p^{-2}}{-2} \\[5ex] = -\dfrac{5}{2} * p^{-2} \\[5ex] = -\dfrac{5}{2} * \dfrac{1}{p^2} \\[5ex] = -\dfrac{5}{2p^2} \\[5ex] = -\dfrac{5}{2(x + 1)^2} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \left[-\dfrac{5}{2(x + 1)^2}\right]_1^2 \\[5ex] = -\dfrac{5}{2(2 + 1)^2} - -\dfrac{5}{2(1 + 1)^2} \\[5ex] = -\dfrac{5}{2(3)^2} + \dfrac{5}{2(2)^2} \\[5ex] = -\dfrac{5}{2(9)} + \dfrac{5}{2(4)} \\[5ex] = -\dfrac{5}{18} + \dfrac{5}{8} \\[5ex] = \dfrac{-20 + 45}{72} \\[5ex] = \dfrac{25}{72} \\[7ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \dfrac{1}{6} + \dfrac{25}{72} \\[5ex] = \dfrac{12 + 25}{72} \\[5ex] = \dfrac{37}{72} \\[5ex] \therefore \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx = \dfrac{37}{72}$
(6.) KCSE A particle was moving along a straight line.
The acceleration of the particle after t seconds was given by $(4t - 13)\;ms^{-2}$.
The initial velocity of the particle was 18 $ms^{-1}$
(a) Determine the value of t when the particle is momentarily at rest.
(b) Find the distance covered by the particle between the time t = 1 second and t = 3 seconds.

$a = 4t - 13 \\[3ex] a = \dfrac{dv}{dt} \\[5ex] dv = a dt \\[3ex] v = \displaystyle\int a dt \\[3ex] v = \displaystyle\int (4t - 13) dt \\[3ex] v = \dfrac{4t^2}{2} - 13t + C \\[3ex] v = 2t^2 - 13t + C \\[3ex] when\;\;t = 0,\;\;initial\;\;velocity = v = 18\;ms^{-1} \\[3ex] \implies \\[3ex] 18 = 2(0)^2 - 13(0) + C \\[3ex] 18 = 0 - 0 + C \\[3ex] 18 = C \\[3ex] C = 18 \\[3ex] v = 2t^2 - 13t + 18 \\[3ex] (a) \\[3ex] momentarily\;\;at\;\;rest\implies v = 0 \\[3ex] 0 = 2t^2 - 13t + 18 \\[3ex] 2t^2 - 13t + 18 = 0 \\[3ex] 2t(t - 2) - 9(t - 2) = 0 \\[3ex] (t - 2)(2t - 9) = 0 \\[3ex] t - 2 = 0 \;\;\;OR\;\;\; 2t - 9 = 0 \\[3ex] t = 2 \;\;\;OR\;\;\; 2t = 9 \\[3ex] t = 2\;s \;\;\;OR\;\;\; t = \dfrac{9}{2}\;s \\[3ex] (b) \\[3ex] v = \dfrac{dd}{dt} \\[5ex] dd = vdt \\[3ex] d = \displaystyle\int v dt \\[3ex]$ The zeros of the function...where the graph include areas above or below the x-axis (t-axis) are t = 2 seconds and t = $\dfrac{9}{2} = 4.5$ seconds
But we are asked to calculate the area (distance covered) between t = 1 second and t = 3 seconds
So, we have to first calculate the area between t = 1 second and t = 2 seconds
Then, we calculate the area between t = 2 seconds and t = 3 seconds
Because 4.5 seconds is greater than 3 seconds, we skip that area.

$v = \displaystyle\int_1^2 (2t^2 - 13t + 18) dt + \displaystyle\int_2^3 (2t^2 - 13t + 18) dt \\[5ex] v = \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 + \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 \\[5ex] t = 2\;s \\[3ex] \dfrac{2(2)^3}{3} - \dfrac{13(2)^2}{2} + 18(2) \\[5ex] \dfrac{16}{3} - \dfrac{26}{1} + \dfrac{36}{1} \\[5ex] \dfrac{16 - 78 + 108}{3} \\[5ex] \dfrac{46}{3}\;m \\[5ex] t = 1\;s \\[3ex] \dfrac{2(1)^3}{3} - \dfrac{13(1)^2}{2} + 18(1) \\[5ex] \dfrac{2}{3} - \dfrac{13}{2} + \dfrac{18}{1} \\[5ex] \dfrac{4 - 39 + 108}{6} \\[5ex] \dfrac{73}{6}\;m \\[5ex] Area = \dfrac{46}{3} - \dfrac{73}{6} = \dfrac{92 - 73}{6} = \dfrac{19}{6}\;m \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] t = 3\;s \\[3ex] \dfrac{2(3)^3}{3} - \dfrac{13(3)^2}{2} + 18(3) \\[5ex] \dfrac{18}{1} - \dfrac{117}{2} + \dfrac{54}{1} \\[5ex] \dfrac{36 - 117 + 108}{2} \\[5ex] \dfrac{27}{2}\;m \\[5ex] t = 2\;s \\[3ex] \dfrac{46}{3}\;m \\[5ex] Area = \dfrac{27}{2} - \dfrac{46}{3} = \dfrac{81 - 92}{6} = -\dfrac{11}{6} \\[5ex] Because\;\;Area\;\;cannot\;\;be\;\;negative; \;\;Area = \dfrac{11}{6}\;m \\[5ex] Total\;\;Area = Distance\;\;covered \\[3ex] = \dfrac{19}{6} + \dfrac{11}{6} \\[5ex] = \dfrac{30}{6} \\[5ex] = 5\;m$
(7.) WASSCE:FM Evaluate $\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx$

$A.\:\: 1\dfrac{1}{3} \\[5ex] B.\:\: 1\dfrac{1}{2} \\[5ex] C.\:\: 3 \\[3ex] D.\:\: 4\dfrac{1}{2}$

Power Rule

$\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx \\[5ex] 3 - 3x^2 = 3(1 - x^2) \\[3ex] 1 - x^2 = 1^2 - x^2 = (1 + x)(1 - x)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] 3 - 3x^2 = 3(1 + x)(1 - x) = 3(x + 1)(1 - x) \\[3ex] \rightarrow \\[3ex] \dfrac{3 - 3x^2}{x + 1} = \dfrac{3(x + 1)(1 - x)}{x + 1} = 3(1 - x) = 3 - 3x \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3 - 3x)dx = \displaystyle\int 3dx - \displaystyle\int 3xdx \\[5ex] \displaystyle\int 3dx = 3x \\[5ex] \displaystyle\int 3xdx = 3 * \dfrac{x^{1 + 1}}{1 + 1} = \dfrac{3x^2}{2} \\[5ex] \rightarrow \\[3ex] \displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx = \left[3x - \dfrac{3x^2}{2}\right]_0^1 \\[5ex] = \left[3(1) - \dfrac{3(1)^2}{2}\right] - \left[3(0) - \dfrac{3(0)^2}{2}\right] \\[5ex] = \left[3 - \dfrac{3(1)}{2}\right] - \left[0 - \dfrac{3(0)}{2}\right] \\[5ex] = \left(3 - \dfrac{3}{2}\right) - \left(0 - \dfrac{0}{2}\right) \\[5ex] = \left(\dfrac{6}{2} - \dfrac{3}{2}\right) - (0 - 0) \\[5ex] = \dfrac{6 - 3}{2} - 0 \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2}$
(8.) Evaluate $\displaystyle\int (3^p + 7^p)dp$

Standard Integral

$\displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[5ex] \displaystyle\int (3^p + 7^p)dp = \displaystyle\int 3^pdp + \displaystyle\int 7^pdp \\[5ex] \displaystyle\int 3^p dx = \dfrac{3^p}{\ln 3} \\[5ex] \displaystyle\int 7^p dx = \dfrac{7^p}{\ln 7} \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3^p + 7^p)dp = \dfrac{3^p}{\ln 3} + \dfrac{7^p}{\ln 7} + C$
(9.) Evaluate $\displaystyle\int \sqrt{2 + \sqrt{x}} dx$

Integration by Algebraic Substitution

$\displaystyle\int \sqrt{2 + \sqrt{x}} dx \\[3ex] Let\:\: p = 2 + \sqrt{x} \\[3ex] \dfrac{dp}{dx} = \dfrac{1}{2\sqrt{x}} \\[5ex] \dfrac{dx}{dp} = 2\sqrt{x} \\[5ex] dx = 2\sqrt{x}dp \\[3ex] \rightarrow \\[3ex] \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \displaystyle\int \sqrt{p} * 2\sqrt{x} dp \\[3ex] = 2\displaystyle\int \sqrt{p} * \sqrt{x} dp \\[3ex] p = 2 + \sqrt{x} \rightarrow \sqrt{x} = p - 2 \\[3ex] = 2 * \displaystyle\int \sqrt{p}(p - 2) dp \\[3ex] \displaystyle\int \sqrt{p}(p - 2) dp = \displaystyle\int p\sqrt{p}dp - \displaystyle\int 2\sqrt{p}dp \\[3ex] \displaystyle\int p\sqrt{p}dp = \displaystyle\int p^1 * p^{\dfrac{1}{2}}dp = \displaystyle\int p^{1 + \dfrac{1}{2}}dp = \displaystyle\int p^{\dfrac{3}{2}}dp \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{p^{\dfrac{3}{2} + 1}}{\dfrac{3}{2} + 1} \\[7ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} \div \dfrac{5}{2} \\[5ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} * \dfrac{2}{5} \\[5ex] p^{\dfrac{5}{2}} = (\sqrt{p})^5 = (\sqrt{p})^4 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^4 * \sqrt{p} = p^2\sqrt{p} \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{2p^2}{5} \\[5ex] \displaystyle\int 2\sqrt{p}dp = 2\displaystyle\int \sqrt{p}dp \\[3ex] \displaystyle\int \sqrt{p}dp = \dfrac{p^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \\[7ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} \div \dfrac{3}{2} \\[5ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} * \dfrac{2}{3} \\[5ex] p^{\dfrac{3}{2}} = (\sqrt{p})^3 = (\sqrt{p})^2 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^2 * \sqrt{p} = p\sqrt{p} \\[5ex] \displaystyle\int 2\sqrt{p}dp = \dfrac{2p\sqrt{p}}{3} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{p}(p - 2) dp = \dfrac{2p^2\sqrt{p}}{5} - \dfrac{2p\sqrt{p}}{3} \\[5ex] = 2p\sqrt{p}\left(\dfrac{p}{5} - \dfrac{1}{3}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p}{15} - \dfrac{5}{15}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] \rightarrow \\[3ex] 2 * \displaystyle\int \sqrt{p}(p - 2) dp = 2 * 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] = 4p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] Substitute\:\:back \\[3ex] = 4(2 + \sqrt{x})\sqrt{2 + \sqrt{x}}\left(\dfrac{3(2 + \sqrt{x}) - 5}{15}\right) \\[5ex] \dfrac{3(2 + \sqrt{x}) - 5}{15} = \dfrac{6 + 3\sqrt{x} - 5}{15} = \dfrac{1 + 3\sqrt{x}}{15} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{2 + \sqrt{x}} dx = 4(2 + \sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right)\left(\dfrac{1 + 3\sqrt{x}}{15}\right) \\[5ex] \therefore \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \dfrac{4}{15}(2 + \sqrt{x})(1 + 3\sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right) + C$
(10.) Find the integral of $\tan x$

Integration by Algebraic Substitution

$\displaystyle\int \tan xdx \\[5ex] \tan x = \dfrac{\sin x}{\cos x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \cos x \\[3ex] \dfrac{dp}{dx} = -\sin x \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{\sin x} \\[5ex] dx = -\dfrac{dp}{\sin x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \tan xdx = \displaystyle\int \dfrac{\sin x}{\cos x} dx \\[5ex] = \displaystyle\int \dfrac{\sin x}{p} * -\dfrac{dp}{\sin x} \\[5ex] = \displaystyle\int -\dfrac{dp}{p} \\[5ex] = -\ln p + C \\[3ex] = -\ln \cos x + C$
(11.)

(12.) Evaluate $\displaystyle\int (v^{-2} + 8v^{-1})dv$

Power Rule

$\displaystyle\int (v^{-2} + 8v^{-1})dv \\[5ex] = \displaystyle\int v^{-2}dv + \displaystyle\int 8v^{-1}dv \\[5ex] = \dfrac{v^{-2 + 1}}{-2 + 1} + 8 * \displaystyle\int v^{-1}dv \\[5ex] = \dfrac{v^{-1}}{-1} + 8 * \ln v + C \\[5ex] = -v^{-1} + 8\ln v + C \\[5ex] = -\dfrac{1}{v} + 8\ln v + C \\[5ex] = 8\ln v - \dfrac{1}{v} + C$
(13.)

(14.)

(15.)

(16.)