\begin{align*} {d \over dx} [ \ln (2x^2 - 3) ] & = {4x \over 2x^2 - 3} \phantom{000000} \left[ {d \over dx} [ \ln f(x) ] = {f'(x) \over f(x)} \right] \end{align*}

\begin{align*} {d \over dx} [ \ln (2x^2 - 3) ] & = {4x \over 2x^2 - 3} \\ \\ \implies \int {4x \over 2x^2 - 3} \phantom{.} dx & = \ln (2x^2 - 3) \\ \\ \therefore \int_2^3 {4x \over 2x^2 - 3} \phantom{.} dx & = \{ \ln [2(3)^2 - 3] \} - \{ \ln [2(2)^2 - 3] \} \\ & = \ln 15 - \ln 5 \\ & = \ln {15 \over 5} \phantom{000000} [\text{Quotient law (logarithms)}] \\ & = \ln 3 \\ \\ \therefore a & = 3 \end{align*}

\begin{align*} u & = x &&& v & = (x + 1)^2 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 2(x + 1) (1) \phantom{000000} [\text{Chain rule}] \\ & &&& & = 2(x + 1) \end{align*} \begin{align*} {d \over dx} \left[ x \over (x + 1)^2 \right] & = { (x + 1)^2 (1) - (x)[2(x + 1)] \over [(x + 1)^2]^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { (x + 1)^2 - 2x (x + 1) \over (x + 1)^4 } \\ & = { (x + 1) [ (x + 1) - 2x ] \over (x + 1)^4 } \phantom{00000000000}[\text{Factorise } (x + 1)] \\ & = { (x + 1) (1 - x) \over (x + 1)^4 } \\ & = { 1 - x \over (x + 1)^3 } \end{align*}

\begin{align*} {d \over dx} \left[ x \over (x + 1)^2 \right] & = { 1 - x \over (x + 1)^3 } \\ \\ \implies \int {1 - x \over (x + 1)^3 } \phantom{.} dx & = {x \over (x + 1)^2 } \\ \\ \\ \int {3 - 2x \over (x + 1)^3 } \phantom{.} dx & = \int { 2 - 2x \over (x + 1)^3 } + {1 \over (x + 1)^3} \phantom{.} dx \\ & = \int { 2 - 2x \over (x + 1)^3 } \phantom{.} dx + \int {1 \over (x + 1)^3 } \phantom{.} dx \\ & = 2 \underbrace{ \int {1 - x \over (x + 1)^3 } \phantom{.} dx }_\text{Match result in line 2} + \int (x + 1)^{-3} \phantom{.} dx \\ & = 2 \left[ x \over (x + 1)^2 \right] + \int (x + 1)^{-3} \phantom{.} dx \\ & = {2x \over (x + 1)^2} + {(x + 1)^{-2} \over (-2)(1)} \phantom{000000} \left[ \int [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over (n + 1). f'(x) } \right] \\ & = {2x \over (x + 1)^2} - {1 \over 2(x + 1)^2} \\ & = {4x \over 2(x + 1)^2} - {1 \over 2(x + 1)^2} \\ & = {4x - 1 \over 2(x + 1)^2} + c \end{align*}

\begin{align*} u & = e^{2x} &&& v & = x \\ {du \over dx} & = 2e^{2x} &&& {dv \over dx} & = 1 \end{align*} \begin{align*} {d \over dx} (x e^{2x} ) & = (x)(2e^{2x}) + (e^{2x})(1) \phantom{000000} [\text{Product rule}] \\ & = 2x e^{2x} + e^{2x} \end{align*}

\begin{align*} {d \over dx} (x e^{2x} ) & = 2x e^{2x} + e^{2x} \\ \\ \implies \int 2x e^{2x} + e^{2x} \phantom{.} dx & = x e^{2x} \\ \underbrace{ \int 2x e^{2x} \phantom{.} dx }_\text{Need this} + \int e^{2x} \phantom{.} dx & = x e^{2x} \\ \int 2x e^{2x} \phantom{.} dx & = x e^{2x} - \int e^{2x} \phantom{.} dx \\ \int 2x e^{2x} \phantom{.} dx & = x e^{2x} - { e^{2x} \over 2 } \phantom{000000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \right] \\ 2 \int x e^{2x} \phantom{.} dx & = x e^{2x} - {1 \over 2} e^{2x} \\ \int x e^{2x} \phantom{.} dx & = {1 \over 2} \left( x e^{2x} - {1 \over 2} e^{2x} \right) \\ & = {1 \over 2} x e^{2x} - {1 \over 4} e^{2x} + c \end{align*}