‹ Back to the paper
Solve the following.
Evaluate the following integrals: Evaluate: ∫ ln x dx. Hence, evaluate: ∫…
Evaluate the following integrals:
(a)[2]
Evaluate: $\int \ln x dx$.
(b)[3]
Hence, evaluate: $\int \frac{\ln[\ln(\frac{1+x}{1-x})]}{1-x^2} dx$.
Answer
Answer (a)
Official answer keyIntegrating by parts with $u = \ln x$ and $dv = dx$: $\int \ln x\,dx = x\ln x - \int \frac{1}{x}\cdot x\,dx = x\ln x - x + c$.
Final answer: $x(\ln x - 1) + c$
Answer (b)
Official answer keyLet $t = \ln\frac{1+x}{1-x}$. Then $dt = \left(\frac{1}{1+x} + \frac{1}{1-x}\right)dx = \frac{2}{1-x^2}dx$.
$\int \frac{\ln\left[\ln\frac{1+x}{1-x}\right]}{1-x^2}dx = \frac{1}{2}\int \ln t\,dt = \frac{1}{2}t(\ln t - 1) + c$ (using part (a)).
Final answer: $\frac{1}{2}\ln\frac{1+x}{1-x}\left[\ln\left(\ln\frac{1+x}{1-x}\right) - 1\right] + c$