\begin{align*} [f(x) = x^2 + 2x - 5, & f'(x) = 2x + 2] \\ \\ \int {x + 1 \over (x^2 + 2x - 5)^2} \times {2 \over 2} \phantom{.} dx & = {1 \over 2} \int {2x + 2 \over (x^2 + 2x - 5)^2} \phantom{.} dx \\ & = {1 \over 2} \int (2x + 2) (x^2 + 2x - 5)^{-2} \phantom{.} dx \\ & = {1 \over 2} \left[ (x^2 + 2x - 5)^{-1} \over -1 \right] \\ & = -{1 \over 2(x^2 + 2x - 5)} + C \end{align*}

\begin{align*} [f(x) = x^2 + 2x - 5, & f'(x) = 2x + 2] \\ \\ \int {4x + 4 \over x^2 + 2x - 5} \phantom{.} dx & = {2 \over 1} \int {2x + 2 \over x^2 + 2x - 5} \phantom{.} dx \\ & = 2 \ln |x^2 + 2x - 5| + C \end{align*}

\begin{align*} [f(x) = \cos x, & f'(x) =- \sin x] \\ \\ \int \sin x \phantom{.} e^{\cos x} \phantom{.} dx & = - \int - \sin x \phantom{.} e^{\cos x} \phantom{.} dx \\ & = -e^{\cos x} + C \end{align*}

\begin{align*} [f(x) = 1 - e^{3x}, & f'(x) =0 - 3e^{3x} = - 3e^{3x}] \\ \\ \int {e^{3x} \over 1 - e^{3x} } \times {-3 \over -3} \phantom{.} dx & = -{1 \over 3} \int { -3e^{3x} \over 1 - e^{3x} } \phantom{.} dx \phantom{00000} \left[ \int {f'(x) \over f(x)} \phantom{.} dx \right] \\ & = -{1 \over 3} \ln |1 - e^{3x}| + C \end{align*}

\begin{align*} & [ f(x) = \sin 3x, f'(x) = 3 \cos 3x] \\ \\ \int \cos 3x \sin^3 3x \phantom{.} dx & = {1 \over 3} \int 3 \cos 3x (\sin 3x)^3 \phantom{.} dx \phantom{000000} \left[ \int f'(x) [f(x)]^n \phantom{.} dx \right] \\ & = {1 \over 3} \left[ (\sin 3x)^4 \over 4 \right] \\ & = {1 \over 12} \sin^4 3x + C \end{align*}

\begin{align*} \int \cot {1 \over 2}x \sec^2 {1 \over 2}x \phantom{.} dx & = \int {1 \over \tan {1 \over 2}x} . \sec^2 {1 \over 2}x \phantom{.} dx \\ & = \int { \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx \\ & = 2 \int { {1 \over 2} \sec^2 {1 \over 2}x \over \tan {1 \over 2}x } \phantom{.} dx \phantom{0000} [ f(x) = \tan {1 \over 2}x, f'(x) = {1 \over 2} \sec^2 {1 \over 2}x ] \\ & = 2 \ln \left| \tan {1 \over 2} x \right| + C \end{align*}

\begin{align*} \int \sin 3x \cos 3x \phantom{.} dx & = {1 \over 2} \int 2 \sin 3x \cos 3x \phantom{.} dx \\ & = {1 \over 2} \int \sin 6x \phantom{.} dx \phantom{00000000} [\sin 2A = 2 \sin A \cos A] \\ & = {1 \over 12} \int 6 \sin 6x \phantom{.} dx \phantom{00000} \left[ \int f'(x) \sin [f(x)] \phantom{.} dx \right]\\ & = {1 \over 12} (-\cos 6x) \\ & = -{1 \over 12} \cos 6x + C \end{align*}

\begin{align*} \text{Since } \cos 2A & = 1 - 2 \sin^2 A, \\ 2 \sin^2 A & = 1 - \cos 2A \\ \sin^2 A & = {1 \over 2} - {1 \over 2} \cos 2A \\ \\ \int \sin^2 2x \phantom{.} dx & = \int \left( {1 \over 2} - {1 \over 2} \cos 4x \right) \phantom{.} dx \\ & = \int {1 \over 2} \phantom{.} dx - {1 \over 2} \int \cos 4x \phantom{.} dx \\ & = {1 \over 2}x - {1 \over 8} \int 4 \cos 4x \phantom{.} dx \\ & = {1 \over 2}x - {1 \over 8} \sin 4x + C \end{align*}

\begin{align*} \text{Since } \cos 2A & = 2 \cos^2 A - 1, \\ - 2 \cos^2 A & = - \cos 2A - 1 \\ 2 \cos^2 A & = \cos 2A + 1 \\ \\ \int 2 \cos^2 \left(x \over 3\right) \phantom{.} dx & = \int \cos \left({2 \over 3}x\right) + 1 \phantom{.} dx \\ & = \int \cos \left( {2 \over 3}x \right) \phantom{.} dx + \int 1 \phantom{.} dx \\ & = {3 \over 2} \int {2 \over 3} \cos \left( {2 \over 3}x \right) \phantom{.} dx + x \\ & = {3 \over 2} \sin \left( {2 \over 3}x \right) + x + C \end{align*}

\begin{align*} \text{Since } \sec^2 A & = 1 + \tan^2 A, \\ \tan^2 A & = \sec^2 A - 1 \\ \\ \int 2 \tan^2 {x \over 3} \phantom{.} dx & = \int 2 \left( \sec^2 {x \over 3} - 1 \right) \phantom{.} dx \\ & = \int 2 \sec^2 {x \over 3} \phantom{.} dx - \int 2 \phantom{.} dx \\ & = 6 \int {1 \over 3} \sec^2 {x \over 3} \phantom{.} dx - 2x \\ & = 6 \tan {x \over 3} - 2x + C \end{align*}

\begin{align*} \text{Since } \text{cosec}^2 A & = 1 + \cot^2 A, \\ \cot^2 A & = \text{cosec}^2 A - 1 \\ \\ \int \cot^2 (\pi x) \phantom{.} dx & = \int \text{cosec}^2 (\pi x) - 1 \phantom{.} dx \\ & = \int \text{cosec}^2 (\pi x) \phantom{.} dx - \int 1 \phantom{.} dx \\ & = {1 \over \pi} \int \pi \text{ cosec}^2 (\pi x) \phantom{.} dx - x \\ & = {1 \over \pi} [ -\cot (\pi x) ] - x \\ & = -{1 \over \pi} \cot (\pi x) - x + C \end{align*}

\begin{align*} \int {1 \over \sqrt{3 - 25x^2} } \phantom{.} dx & = \int { 1 \over \sqrt{25} \sqrt{ {3 \over 25} - x^2 } } \phantom{.} dx \phantom{000000} [\text{Ensure coefficient of } x^2 \text{ is 1}] \\ & = {1 \over 5} \int { 1 \over \sqrt{ \left(\sqrt{3} \over 5\right)^2 - x^2 } } \phantom{.} dx \\ & = {1 \over 5} \sin^{-1} \left( x \over {\sqrt{3} \over 5} \right) \\ & = {1 \over 5} \sin^{-1} \left(5x \over \sqrt{3} \right) + C \end{align*}

\begin{align*} x^2 - 3x + 2 & = x^2 - 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 + 2 \\ & = \left(x - {3 \over 2}\right)^2 - {1 \over 4} \phantom{00000000000000} [\text{Complete the square}] \\ \\ \int {1 \over x^2 - 3x + 2} \phantom{.} dx & = \int {1 \over \left(x - {3 \over 2}\right)^2 - {1 \over 4}} \phantom{.} dx \\ & = \int {1 \over \left(x - {3 \over 2}\right)^2 - \left(1 \over 2\right)^2} \phantom{.} dx \\ & = {1 \over 2\left(1 \over 2\right)} \ln \left| x - {3 \over 2} - {1 \over 2} \over x - {3 \over 2} + {1 \over 2} \right| \\ & = \ln \left| {x - 2 \over x - 1} \right| + C \end{align*}

\begin{align*} {1 \over x^2 - 3x + 2} & = {1 \over (x - 1)(x - 2)} \\ & = {A \over x - 1} + {B \over x - 2} \\ & = {A (x - 2) + B (x - 1) \over (x - 1)(x - 2)} \\ \\ 1 & = A(x - 2) + B(x - 1) \\ \\ \text{Let } & x = 2, \\ 1 & = B \\ \\ \text{Let } & x = 1, \\ 1 & = -A \\ -1 & = A \\ \\ {1 \over x^2 - 3x + 2} & = {-1 \over x - 1} + {1 \over x - 2} \\ \\ \int {1 \over x^2 - 3x + 2} \phantom{.} dx & = \int \left( {-1 \over x - 1} + {1 \over x - 2} \right) \phantom{.} dx \\ & = - \int {1 \over x - 1} \phantom{.} dx + \int {1 \over x - 2} \phantom{.} dx \\ & = - \ln |x - 1| + \ln |x - 2| \\ & = \ln \left| {x - 2 \over x - 1} \right| + C \end{align*}

\begin{align*} f(x) & = 5 - 4x - x^2, f'(x) = -4 - 2x \\ \\ \int { 2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = - \int {-2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\ & = - \int \left( {- 2x - 4 \over \sqrt{5 - 4x - x^2} } + {4 \over \sqrt{5 - 4x - x^2} } \right) \phantom{.} dx \\ & = - \int {-2x - 4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx - \int {4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\ & = - \int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx - 4 \int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\ \\ \int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx & = {(5 - 4x - x^2)^{1 \over 2} \over {1 \over 2}} \\ & = 2 \sqrt{ 5 - 4x - x^2 } \\ \\ 5 - 4x - x^2 & = -x^2 - 4x + 5 \\ & = -(x^2 + 4x) + 5 \\ & = - [ (x + 2)^2 - (2)^2 ] + 5 \\ & = -(x + 2)^2 + 4 + 5 \\ & = -(x + 2)^2 + 9 \\ & = 9 - (x + 2)^2 \\ & = 3^2 - (x + 2)^2 \\ \\ \int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = \int {1 \over \sqrt{3^2 - (x + 2)^2}} \phantom{.} dx \\ & = \sin^{-1} \left(x + 2 \over 3\right) \\ \\ \therefore \text{Required answer} & = - 2 \sqrt{ 5 - 4x - x^2 } - 4 \sin^{-1} \left(x + 2 \over 3\right) + C \end{align*}

\begin{align*} f(x) & = x^2 + 3x + 5, f'(x) = 2x + 3 \\ \\ \int {2x^2 \over x^2 + 3x + 5} \phantom{.} dx & = \int 2 + {-6x - 10 \over x^2 + 3x + 5} \phantom{.} dx \\ & = \int 2 \phantom{.} dx - \int {6x + 10 \over x^2 + 3x + 5} \phantom{.} dx \\ & = 2x - \int \left( {6x + 9 \over x^2 + 3x + 5} + {1 \over x^2 + 3x + 5} \right) \phantom{.} dx \\ & = 2x - \int {6x + 9 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over x^2 + 3x + 5} \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over\left(x + {3 \over 2}\right)^2 - \left(3 \over 2\right)^2 + 5} \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + {11 \over 4} } \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + \left(\sqrt{11} \over 2\right)^2 } \phantom{.} dx \\ & = 2x - 3 \ln (x^2 + 3x + 5) - {1 \over {\sqrt{11} \over 2}} \tan^{-1} \left(x + {3 \over 2} \over {\sqrt{11} \over 2} \right) \\ & = 2x - 3 \ln (x^2 + 3x + 5) - {2 \over \sqrt{11}} \tan^{-1} \left(2x + 3 \over \sqrt{11}\right) + C \end{align*}