Methods of Integration

$\int \frac{1}{x^2\:+\:8x\:+\:25}\,dx$

$= \frac{1}{3}\int \frac{3}{(x+4)^2+3^2}\:dx$

$= \frac{1}{3}\:tan^{-1}\left (\frac{x\:+\:4}{3}\right ) + C$

$\int \frac{4x\:+\:5}{x^2\:+\:2x\:+\:2}\:dx$

$= \int \left(\frac{2(2x\:+\:2)}{x^2\:+\:2x\:+\:2} \:+\:\frac{1}{(x\:+\:1)^2\:+\:1}\right)$

$2\:\Ln(x^2\:+\:2x\:+\:2)\:+\:\tan^{-1}(x\:+\:1) + C$

$\int \frac{1}{\sqrt{x^2+2x-3}}\; dx$

$= \int \frac{1}{\sqrt[]{(x+1)^2-4}}$ $=cosh^{-1}(\frac{x+1}{2}) + C$

Other forms $\int \frac{b}{(x + a)^2-d^2}\:dx = b \cosh^{-1} \frac{x + a}{d}$ $\int \frac{b}{\sqrt[]{(x+a)^2+d^2}} = b \sinh^{-1} \frac{x + a}{d}$ $\int \frac{b}{\sqrt[]{d^2-(x+a)^2}} = b \sin^{-1} \frac{x + c}{d}$

$\int \frac{x\:+\:1}{\sqrt[]{x^2\:+\:2x\:+\:3}}\:dx\:= \sqrt[]{x^2\:+\:2x\:+\:3} + C$

but $\frac{d\:\sqrt[]{x^2\:+\:2x\:+\:3}}{dx} = \frac{1}{2}(x^2\:+\:2x\:+\:3)^{-\frac{1}{2}}\:X\:(2x\:+\:3)$ $=\:\frac{\frac{1}{2}(2x\:+\:2)}{\sqrt[]{x^2\:+\:2x\:+\:3}}$

From which it can be seen that $\frac{d\int_{}^{}f\:x}{dx} = \frac{\frac{1}{2}f^{'}x}{\sqrt[]{fx}}$

$\int \frac{3x\:-\:5}{x^2\:+\:4x\:+\:7}\;dx$

$\int \left(\frac{3\:.\:\frac{1}{2}(2x\:+\:4)}{\sqrt[]{x^2\:+\:4x\:+\:7}} \,-\:\frac{11}{\sqrt[]{(x^2\:+\:2)^2}\:+\:3}}\right)\:dx$ $= 3\:\sqrt[]{x^2\:+\:4x\:+\:7}\:-\:11\:sinh^{-1}\frac{x\:+\:2}{\sqrt[]{3}} + C$

In general the answer to this type are in the form $\sqrt[]{} + (\sinh\:or\:\cosh\:or\:\sin)^{-1}$ The integral of $\sqrt[]{\frac{x\:-\:1}{2x\:+\:1}}\;dx$ can be found by multiplying top and bottom by $\sqrt[]{x\:-\:1}$

thus $\int \sqrt[]{\frac{x\:-\:1}{2x\:+\:3}} = \int\frac{x\:-\:1}{\sqrt[]{2x^2\:+\:x\:-\:3}}$ $=\:\frac{1}{2}\sqrt[]{2x^2\:+\:x\:-\:3} - \frac{5}{4\;\sqrt[]{2}}\cosh^{-1}\frac{4x\:+\:1}{5} + C$

Find the integral of $\int \frac{1}{x+\sqrt{2x - 1}}\;dx$

Substitute $U=\sqrt{2x - 1}$ i.e. $x=\frac{U^2+1}{2}$

Therefore $U\:dU = dx$ \todo Review the following workings

thus the integral can be written as:- $\int \frac{1}{\frac{U^2+1}{2}+U}\:dU$$= 2\,\int \frac{U}{(U^2+1)}dU$ $= 2\int\left(\frac{1}{U+1}-\frac{1}{(U+1)^2} \right)$ $= 2\ln (U+1)+\frac{2}{U+1}+C$

Therefore $\int \frac{1}{x+\sqrt{2x-1}}\:dx = 2 \ln \left\{ \sqrt{2x-1} +1 \right\} +\frac{2}{\sqrt{2x-1} + 1}+ C$

To find the integral of $(\cos \:5x\:\cos\:2x)\;dx$

But $\cos\:A\:+\:B = 2\:\cos\frac{A\:+\:B}{2}\:\cos\frac{A\,-\:B}{2}$ from which it can be shown that $\int\cos\,5x\:\cos\,2x\:dx = \frac{1}{2}\int (\cos\,7x\:+\:\cos\,3x)\:dx$

$=\:\frac{1}{14}\,\sin\,7x\:+\:\frac{1}{6}\sin\,3x + C$

$\int\ cos^2\:x\:dx = \frac{1}{2}\:\int (2\:\cos\,2x\:+\:1)$

$= \frac{\sin\:2x}{4}\:+\:\frac{x}{2}\:+\:C$

$\int \cosec\:x\:dx = \int \frac{1}{sin\,x}\:dx$

$= \int \frac{1}{\frac{2t}{1\:+\:t^2}}\;X\;\frac{2}{1\:+\:t^2}\;dt$ $= \int \frac{1}{t}\:dt = \ln\:t + C$

Therefore $\int \cosec\:x\:dx = \ln\,\tan\frac{x}{2} + C$

$\int \sec\:x\;dx\;=\:\int \frac{1}{\cos\:x}\:dx$

using the same substitution as above $=\:\int \frac{1\:+\:t^2}{1\:-\:t^2}\;.\;\frac{2}{1\:+\:t^2}\:dt$ $=\:\int \left(\frac{1}{1\:-\:t}\:+\:\frac{1}{1\:-\:t} \right)\:dt$ $=\:-\:\Ln\,(1\:-\:t)\:+\:\Ln\,(1\:+\:t)\:+\:C$ $=\:Ln\:\frac{1\:+\:t}{1\:-\:t}\;+\:C$

Therefore $\int \sec\:x\;dx\:=$ $\:\Ln\frac{1\:+\:tan\,\frac{x}{2}}{1\:-\:tan\frac{x}{2}} + C$

$\int \frac{1}{x^2\:\sqrt[]{1\:-\:x^2}}\:dx$

$put\;\;\;\;x\:=\:\sin\:\theta\;\;\:and\;\therefore\;\;\;dx\:=\:\cos\:\theta\:d\theta$ $thus\;\;\;integral\:=\:\int \frac{1}{sin^2\theta\:\cos\,\theta}\:\cos\theta\:d\theta$ $=\:\int cosec^2\:\theta\:d\theta$

$=\:\cot\:\theta\:+\:C\;\;=\:-\:\frac{\sqrt[]{1\:-\:x^2}}{x}\:+\:C$

Find the integral of $\sqrt[]{(x^2\:+\:a^2)}\;dx$

Let $x = a sinh u$ $\therefore\;\;\;\frac{dy}{dx}\:=\:a\:cosh\,u$ $\frac{dy}{du}\:=\:\frac{dy}{dx}\:X\:\frac{dx}{du}\:=\sqrt[]{(x^2\:+\:a^2)}\:X\:a\:cosh\,u$ $but\;\;\;\sqrt[]{(x^2\:+\:a^2}\:=\:\sqrt[]{a^2\,\sinh^2\,u\:+\:a^2)}=\:\sqrt[]{a^2\:\cosh^2\,u}$ $thus\;\;\;\frac{dy}{du}\:=\:a^2\:\cosh^2\,u\:=\:\frac{1}{2}\,a^2(1\:+\:\cosh\,2u)$ $\therefore\;\;\;y\:=\:\frac{1}{2}\:a^2\int (1\:+\:\cosh\:2u)\:du$ $=\:\frac{1}{2}\:a^2\:(u\:+\:\frac{1}{2}\,\sinh\,2u)\:=\frac{1}{2}\:a^2\,u\:+\:\frac{1}\:(sinh\,u\:\cosh\,u)$

$=\:\frac{1}{2}\:a^2\:\sinh^{-1}\frac{x}{a}\:+\:x\:\sqrt[]{(a^2\:+\:x^2)}$

$\int x \cdot cos x\:dx$

$=\int x \cdot (\sin x)'\:dx = x \sin x-\int \sin x\:dx$

$=x \sin x + \cos x + C$

$\int x^2\ln x\;dx$

Can be written as:- $\int \ln x\left ( \frac{1}{3}x^3 \right )'\;dx=\frac{1}{3}\ln x-\int \frac{1}{3}x^3\times \frac{1}{x}\;dx$

$=\frac{1}{3}x^3\ln x-\frac{1}{9}x^3+C$