Question

solve $\int [1+\frac{1}{(1+x^2)}-\frac{2}{\sqrt{1-x^2}}+\frac{5}{x\sqrt{x^2-1}}+a^x]dx$

Answer 1

$\int (1+\frac{1}{1+x^2}-\frac{2}{\sqrt{1-x^2}}+\frac{5}{x\sqrt{x^2-1}}+a^x)dx$

$=\int dx+\int \frac{dx}{1+x^2}-2\int \frac{dx}{\sqrt{1-x^2}}+5\int \frac{dx}{x\sqrt{x^2-1}}+\int a^{x}dx$

$=x+{\tan }^{-1}x-2{\sin }^{-1}\frac{x}{a}+5I_1+\int a^{x}dx⟶\left(1\right)$

Where $I_1=\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{-\cos \theta }{{\sin }^2\theta \frac{1}{\sin \theta }\sqrt{\frac{1-{\sin }^2\theta }{\sin \theta }}}d\theta$

Let $x=\frac{1}{\sin \theta }=\int \frac{-\cos \theta }{\cos \theta }d\theta$

$\Rightarrow dx=-{\left(\sin \theta \right)}^{-1-1}\cos \theta =-\theta$

$=\frac{-\cos \theta }{{\sin }^2\theta }d\theta =-{\sin }^2\frac{1}{x}$

From $\left(1\right)$ we have,

$=x+{\tan }^{-1}x-2\sin -x-5{\sin }^{-1}\frac{1}{x}+\frac{ax}{ma}+c$

Hence, the answer is $x+{\tan }^{-1}x-2\sin -x-5{\sin }^{-1}\frac{1}{x}+\frac{ax}{ma}+c.$