Here, we will discuss some standard integrals which are directly related to standard differentiation formulae and so may be considered as some of the fundamental integrals. [Antiderivatives]
$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$
Put $x=a\tan\theta$, then $dx=a\sec^2\theta$. \[\therefore\int\frac{1}{a^2+x^2}dx=\int\frac{1}{a^2+a^2\tan^2\theta}a\sec^2\theta d\theta\] \[=\int\frac{1}{a^2\sec^2\theta}a\sec^2\theta d\theta=\frac{1}{a}\int d\theta\] \[=\frac{1}{a}\theta+C=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\]
$\int\frac{1}{a^2-x^2}dx=\frac{1}{2a}\log\frac{a+x}{a-x}+C$
\[\int\frac{1}{a^2-x^2}dx=\int\frac{1}{(a+x)(a-x)}dx\] \[=\frac{1}{2a}\int\left(\frac{1}{a+x}+\frac{1}{a-x}\right)dx\] \[=\frac{1}{2a}[\log(a+x)-\log(a-x)]+C\]\[=\frac{1}{2a}\log\frac{a+x}{a-x}+C\]
$\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\log\frac{x-a}{x+a}+C$
\[\int\frac{1}{x^2-a^2}dx=\int\frac{1}{(x+a)(x-a)}dx\] \[=\frac{1}{2a}\int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx\] \[=\frac{1}{2a}[\log(x-a)-\log(x+a)]+C\] \[=\frac{1}{2a}\log\frac{x-a}{x+a}+C\]
$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C$
Put $x=a\sin\theta$, then $dx=a\cos\theta d\theta$. \[\therefore\int\frac{dx}{\sqrt{a^2-x^2}}=\int\frac{a\cos\theta d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta}{a\cos\theta}d\theta\] \[=\int d\theta=\theta+C=\sin^{-1}\frac{x}{a}+C\]
$\int\frac{dx}{\sqrt{x^2+a^2}}=\log(x+\sqrt{x^2+a^2})+C$\[=\sinh^{-1}\frac{x}{a}+C\]
Put $x=a\tan\theta$, then, $dx=a\sec^2\theta d\theta$. \[\therefore\int\frac{dx}{\sqrt{x^2+a^2}}=\int\frac{a\sec^2\theta d\theta}{\sqrt{a^2\tan^2\theta+a^2}}\]\[=\int\frac{a\sec^2\theta d\theta}{a\sec\theta}=\int a\sec\theta d\theta\] \[=\log(\sec\theta+\tan\theta)+C’\]\[=\log\left(\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right)+C’\] \[=\log\left(\frac{x+\sqrt{a^2+x^2}}{a}\right)+C’\] \[=\log(x+\sqrt{a^2+x^2})-\log a+C’\] \[=\log(x+\sqrt{a^2+x^2})+C\] Again, put $x=a\sinh y$, then $dx=a\cosh ydy$. \[\therefore\int\frac{dx}{\sqrt{x^2+a^2}}=\int\frac{a\cosh ydy}{\sqrt{a^2\sinh^2y+a^2}}\] \[=\int\frac{a\cosh y}{a\cosh y}dy=\int dy\] \[=y+C=\sinh^{-1}\frac{x}{a}+C\]
$\int\frac{dx}{\sqrt{x^2-a^2}}=\log(x+\sqrt{x^2-a^2})+C$ \[=\cosh^{-1}\frac{x}{a}+C\]
Put $x=a\sec\theta$, then $dx=a\sec\theta\tan\theta d\theta$. \[\therefore\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}}\] \[=\int\frac{a\sec\theta\tan\theta d\theta}{a\tan\theta}=\int\sec\theta d\theta\] \[=\log(\sec\theta+\tan\theta)+C’\]\[=\log\left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right)+C’\] \[=\log\left(\frac{x+\sqrt{x^2-a^2}}{a}\right)+C’\] \[=\log(x+\sqrt{x^2-a^2})-\log a+C’\] \[=\log(x+\sqrt{x^2-a^2})+C\] Again, put $x=a\cosh y$, then, $dx=a\sinh ydy$. \[\therefore\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sinh ydy}{\sqrt{a^2\cosh^2y-a^2}}\] \[=\int\frac{a\sinh y}{a\sinh y}dy=\int dy\] \[=y+C=\cosh^{-1}\frac{x}{a}+C\]
Some Integrals Reducible to Standard Forms
The integrals of the forms $\int\frac{dx}{ax^2+bx+c}$ and $\int\frac{dx}{\sqrt{ax^2+bx+c}}$ can be easily evaluated by converting into above standard integrals. The other forms of integrals which can be reduced to the standard integrals are \[\int\frac{mx+e}{ax^2+bx+c}dx\text{ and }\int\frac{mx+e}{\sqrt{ax^2+bx+c}}dx\] The integral \[\int\frac{mx+e}{ax^2+bx+c}dx\text{ __(1)}\] can be written as \[p\int\frac{2ax+b}{ax^2+bx+c}dx+q\int\frac{1}{ax^2+bx+c}dx\text{ __(2)}\] in which the numerator of the first integral is the differential coefficient of the denominator and the constants $p$ and $q$ are to be adjusted so as to make the sum $\text{(2)}$ equal to the given integral $\text{(2)}$. Here, $2ap=m$ and $bp+q=e$ which can be solved for $p$ and $q$. The integrals in $\text{(2)}$ can be easily integrated.
Similarly, the integral \[\int\frac{mx+e}{\sqrt{ax^2+bx+c}}dx\] can be written as \[p\int\frac{2ax+b}{\sqrt{ax^2+bx+c}}dx+q\int\frac{1}{\sqrt{ax^2+bx+c}}dx\] and the values of $p$ and $q$ be obtained from $2ap=m$ and $bp+q=e$.
$\int\frac{2x+3}{\sqrt{x^2+4x+20}}dx$
\[\text{Let }I=\int\frac{2x+3}{\sqrt{x^2+4x+10}}dx\] \[=\int\frac{(2x+4)-1}{\sqrt{x^2+4x+20}}dx\] \[=\int\frac{2x+4}{\sqrt{x^2+4x+20}}dx-\int\frac{dx}{\sqrt{x^2+4x+20}}\] Put $x^2+4x+20=y$, then, $(2x+4)dx=dy$. \[\therefore I=\frac{dy}{\sqrt{y}}-\int\frac{dx}{\sqrt{x^2+4x+4+16}}\] \[=2\sqrt{y}-\int\frac{dx}{\sqrt{(x+2)^2+4^2}}\] \[=2\sqrt{x^2+4x+20}-\log(x+2+\sqrt{x^2+4x+20})+C\]
$\int\frac{dx}{\sqrt{(x-\alpha)(x-\beta)}}(\beta>\alpha)$
\[\text{Let }I=\int\frac{dx}{\sqrt{(x-\alpha)(x-\beta)}}\] Put $x-\alpha=y^2\Rightarrow x=\alpha+y^2$ so that $dx=2ydy$. \[\therefore I=\int\frac{2ydy}{\sqrt{y^2(\alpha+y^2-\beta)}}=2\int\frac{dy}{\sqrt{y^2-(\beta-\alpha)}}\] \[=2\int\frac{dy}{\sqrt{y^2-(\sqrt{\beta-\alpha})^2}}\] \[=2\log(y+\sqrt{y^2-(\beta-\alpha)})+C\] \[=2\log(\sqrt{x-\alpha}+\sqrt{x-\alpha-\beta+\alpha})+C\] \[=2\log(\sqrt{x-\alpha}+\sqrt{x-\beta})+C\]
$\int\frac{\cos xdx}{\sin^2x+4\sin x+5}$
\[\text{Let }I=\int\frac{\cos xdx}{\sin^2x+4\sin x+5}\] Put $\sin x=y$, then, $\cos xdx=dy$. \[\therefore I=\frac{dy}{y^2+4y+5}\] \[=\int\frac{dy}{y^2+4y+4+1}=\int\frac{dy}{(y+2)^2+1^2}\] \[=\tan^{-1}(y+2)+C\] \[=\tan^{-1}(\sin x+2)+C\]
More Standard Integrals