Substitution Rule for Indefinite Integrals — Question 7

Question 7 - Solution

Let $$u=\sqrt{1-x^2}.$$ Then $$du=\frac{-x}{\sqrt{1-x^2}}\,dx \qquad\Rightarrow\qquad \frac{x}{\sqrt{1-x^2}}\,dx=-du.$$ So the integral becomes $$-\int \ln(1+u)\,du.$$

Now integrate by parts. Take $$w=\ln(1+u), \qquad dv=du,$$ so $$dw=\frac{1}{1+u}\,du, \qquad v=u.$$ Then $$-\int \ln(1+u)\,du =-\left(u\ln(1+u)-\int \frac{u}{1+u}\,du\right).$$

Simplify $$\frac{u}{1+u}=1-\frac{1}{1+u},$$ hence $$\int \frac{u}{1+u}\,du=\int\left(1-\frac{1}{1+u}\right)du =u-\ln(1+u).$$

Therefore, $$-\left(u\ln(1+u)-u+\ln(1+u)\right) = -u\ln(1+u)+u-\ln(1+u)+C.$$

Substitute back $u=\sqrt{1-x^2}$: $$\boxed{ -\sqrt{1-x^2}\,\ln\!\bigl(1+\sqrt{1-x^2}\bigr) +\sqrt{1-x^2} -\ln\!\bigl(1+\sqrt{1-x^2}\bigr) + C }$$