JEE Main & Advanced Mathematics Indefinite Integrals Fundamental Integration Formulae

(1) (i) \[\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}+c,\,n\ne -1}\]

(ii) \[\int{{{(ax+b)}^{n}}dx=\frac{1}{a}.\,\frac{{{(ax+b)}^{n+1}}}{n+1}}+c\], \[n\ne -1\]

(2) (i) \[\int{\frac{1}{x}dx=\log |x|+c}\]

(ii) \[\int{\frac{1}{ax+b}\,dx=\frac{1}{a}(\log |ax+b|)+c}\]

(3) \[\int{{{e}^{x}}dx={{e}^{x}}+c}\]      

(4)  \[\int{{{a}^{x}}dx=\frac{{{a}^{x}}}{{{\log }_{e}}a}+c}\]  

(5)  \[\int{\sin x\,dx=-\cos x+c}\]          

(6)  \[\int{\cos x\,dx=\sin x+c}\]           

(7)  \[\int{{{\sec }^{2}}}x\,dx=\tan x+c\]              

(8)  \[\int{\text{cos}\text{e}{{\text{c}}^{2}}x\,dx}=-\cot x+c\]            

(9) \[\int{\sec x\,\tan x\,dx=\sec x+c}\]               

(10) \[\int{\text{cosec}\,x\,\cot x\,dx=-\text{cosec}\,x+c}\]

(11) \[\int{\tan x\,dx}=-\log |\cos x|+c=\log |\sec x|+c\]

(12) \[\int{\cot x\,dx=\log |\sin x|+}c=-\log |\cos ec\,x|+c\]            

(13) \[\int{\sec x\,dx=\log |\sec x+\tan x|+c=\log \tan \left( \frac{\pi }{4}+\frac{x}{2} \right)+c}\]

(14) \[\int{\text{cos}\text{ec}}\,x\,dx=\log |\text{cos}\text{ec}\,x-\cot x|+c=\log \tan \frac{x}{2}+c\]

(15) \[\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}={{\sin }^{-1}}x+c=-{{\cos }^{-1}}x+c}\]  

(16) \[\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\frac{x}{a}+c}=-{{\cos }^{-1}}\frac{x}{a}+c\]

(17) \[\int{\frac{dx}{1+{{x}^{2}}}={{\tan }^{-1}}x+c=-{{\cot }^{-1}}x+c}\]

(18) \[\int{\frac{dx}{{{a}^{2}}+{{x}^{2}}}=\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}+c=\frac{-1}{a}{{\cot }^{-1}}\frac{x}{a}+c}\]

(19) \[\int{\frac{dx}{x\sqrt{{{x}^{2}}-1}}={{\sec }^{-1}}x+c}=-\cos e{{c}^{-1}}x+c\]            

(20) \[\int{\frac{dx}{x\sqrt{{{x}^{2}}-{{a}^{2}}}}=\frac{1}{a}{{\sec }^{-1}}\frac{x}{a}+c}=\frac{-1}{a}\cos e{{c}^{-1}}\frac{x}{a}+c\]

In any of the fundamental integration formulae, if \[x\] is replaced by \[ax+b\], then the same formulae is applicable but we must divide by coefficient of \[x\] or derivative of \[(ax+b)\] i.e., \[a\]. In general, if \[\int{f(x)dx=\varphi (x)+c}\], then\[\int{f(ax+b)\,dx=\frac{1}{a}\varphi \,(ax+b)}+c\]

\[\int{\sin (ax+b)\,dx=\frac{-1}{a}}\cos (ax+b)+c,\]

\[\int{\sec (ax+b)\,dx=\frac{1}{a}\log |\sec (ax+b)+\tan (ax+b)|+c}\] etc.

Some more results :

(i) \[\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}=\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|+c=\frac{-1}{a}{{\coth }^{-1}}\frac{x}{a}+c}\], when \[x>a\]

(ii) \[\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}dx=\frac{1}{2a}\log \left| \frac{a+x}{a-x} \right|+c=\frac{1}{a}{{\tanh }^{-1}}\frac{x}{a}+c}\], when \[x<a\]

(iii) \[\int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\log \{|x+\sqrt{{{x}^{2}}-{{a}^{2}}}|\}+c=\cos \,{{\text{h}}^{-1}}\left( \frac{x}{a} \right)}+c\]

(iv) \[\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\log }\{|x+\sqrt{{{x}^{2}}+{{a}^{2}}}|\}+c=\sin {{\text{h}}^{-1}}\left( \frac{x}{a} \right)+c\]

(v) \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{1}{2}x\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \frac{x}{a} \right)+c}\]

(vi) \[\int_{{}}^{{}}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2}x\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{1}{2}{{a}^{2}}\log \{x+\sqrt{{{x}^{2}}-{{a}^{2}}}\}+c\]\[=\frac{1}{2}x\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{1}{2}{{a}^{2}}{{\cosh }^{-1}}\left( \frac{x}{a} \right)+c\]

(vii) \[\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\frac{1}{2}x\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{1}{2}{{a}^{2}}\log \{x+\sqrt{{{x}^{2}}+{{a}^{2}}}\}+c}\]\[=\frac{1}{2}x\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{1}{2}{{a}^{2}}\sin {{\text{h}}^{-1}}\left( \frac{x}{a} \right)\]