Substitution Rule for Indefinite Integrals — Question 10

Question 10

Evaluate the integral $$\int x \ln(1-x^2)\,dx.$$

Question 10 - Solution

Let $$u = 1 - x^2.$$ Then $$du = -2x\,dx.$$

Rewrite the integral: $$\int x \ln(1-x^2)\,dx = -\frac12 \int \ln u \, du.$$

Now integrate: $$\int \ln u \, du = u\ln u - u.$$

Therefore, $$-\frac12 (u\ln u - u) + C.$$

Substitute back: $$\boxed{ -\frac12\bigl[(1-x^2)\ln(1-x^2) - (1-x^2)\bigr] + C }$$