We have to do integration with substitution.
I tried to solve it and got stuck after getting $\int \frac{dz}{x(z+1)}$
The answer is given as $2tan^{-1}(x +\sqrt {x^2 +2x-1})$
Integration using fixed method
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integration
1 Answers
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$z-x=\sqrt{x^2+2x-1}$
$z^2+x^2-2zx=x^2+2x-1$
$z^2+1=2zx+2x$
$zx+x=(z+1)x$