$\int \frac{1}{x\left[(\log x)^2+4 \log x-1\right]} d x$ is equal to :
Answer & explanation
Correct answer: option 4
Let $I=\int \frac{1}{x\left[(\ln x)^2+4 \ln x-1\right]} d x$
Let $\ln x=t \Rightarrow \frac{1}{x} dx=dt$
∴ $I=\int \frac{d t}{t^2+4 t -1}=\int \frac{d t}{(t+2)^2-(\sqrt{5})^2}=\frac{1}{2 \sqrt{5}} \ln \left|\frac{t+2-\sqrt{5}}{t+2+\sqrt{5}}\right|+c$
$I=\frac{1}{2 \sqrt{5}} \ln \left|\frac{\ln x+2-\sqrt{5}}{\ln x+2+\sqrt{5}}\right|+c$
Hence (4) is the correct answer.