Question

If $\int \frac{1}{2x^2+x-1}dx=\frac{1}{k}\ln \left|\frac{2x-1}{2x+2}\right|+C$, then the value of $k^2+1$ is
(where $C$ is integration constant)

Answer 1

$I=\int \frac{1}{2x^2+x-1}dx=\frac{1}{2}\int \frac{dx}{x^2+\frac{x}{2}-\frac{1}{2}}=\frac{1}{2}\int \frac{dx}{{(x+\frac{1}{4})}^2-{\left(\frac{3}{4}\right)}^2}=\frac{1}{2}\times \frac{1}{2\cdot \frac{3}{4}}\ln \left|\frac{x+\frac{1}{4}-\frac{3}{4}}{x+\frac{1}{4}+\frac{3}{4}}\right|+C=\frac{1}{3}\ln \left|\frac{2x-1}{2x+2}\right|+C\Rightarrow k=3\therefore k^2+1=10$