Indefinite Integrals — Question 8

Question 8 - Solution

Split the integral: $$\int \ln\!\left(x+\sqrt{1+x^2}\right)\,dx \;+\; \int \frac{x^2}{\sqrt{1+x^2}}\,dx.$$

Part A: $\displaystyle \int \ln\!\left(x+\sqrt{1+x^2}\right)\,dx$

Use integration by parts with $$u=\ln\!\left(x+\sqrt{1+x^2}\right),\qquad dv=dx,\qquad v=x.$$

Let $$g(x)=x+\sqrt{1+x^2}.$$ Then $$du=\frac{g'(x)}{g(x)}\,dx, \qquad g'(x)=1+\frac{x}{\sqrt{1+x^2}}.$$ Thus $$du=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}\,dx =\frac{1}{\sqrt{1+x^2}}\,dx.$$

Therefore, $$\int \ln\!\left(x+\sqrt{1+x^2}\right)\,dx = x\ln\!\left(x+\sqrt{1+x^2}\right) -\int \frac{x}{\sqrt{1+x^2}}\,dx.$$ Let $w=1+x^2$, $dw=2x\,dx$: $$\int \frac{x}{\sqrt{1+x^2}}\,dx =\frac{1}{2}\int w^{-1/2}\,dw =\sqrt{1+x^2}.$$ Hence, $$\int \ln\!\left(x+\sqrt{1+x^2}\right)\,dx = x\ln\!\left(x+\sqrt{1+x^2}\right)-\sqrt{1+x^2}.$$

Part B: $\displaystyle \int \frac{x^2}{\sqrt{1+x^2}}\,dx$ (use an easier trigonometric substitution)

Let $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta, \qquad \sqrt{1+x^2}=\sec\theta.$$ Then $$\int \frac{x^2}{\sqrt{1+x^2}}\,dx = \int \frac{\tan^2\theta}{\sec\theta}\,\sec^2\theta\,d\theta = \int \tan^2\theta\,\sec\theta\,d\theta.$$

Use $\tan^2\theta=\sec^2\theta-1$: $$\int \tan^2\theta\,\sec\theta\,d\theta = \int (\sec^3\theta-\sec\theta)\,d\theta.$$

Evaluate each term directly: $$\int \sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|,$$ and for $\int\sec^3\theta\,d\theta$, integrate by parts: $$\int \sec^3\theta\,d\theta =\frac{1}{2}\left(\sec\theta\tan\theta +\ln|\sec\theta+\tan\theta|\right).$$

Thus, $$\int (\sec^3\theta-\sec\theta)\,d\theta = \frac{1}{2}\sec\theta\tan\theta -\frac{1}{2}\ln|\sec\theta+\tan\theta|.$$

Return to $x$ using $\tan\theta=x$, $\sec\theta=\sqrt{1+x^2}$: $$\int \frac{x^2}{\sqrt{1+x^2}}\,dx = \frac{1}{2}x\sqrt{1+x^2} -\frac{1}{2}\ln\!\left(x+\sqrt{1+x^2}\right).$$

Combine Parts A and B: $$\boxed{ x\ln\!\left(x+\sqrt{1+x^2}\right) -\sqrt{1+x^2} +\frac{1}{2}x\sqrt{1+x^2} -\frac{1}{2}\ln\!\left(x+\sqrt{1+x^2}\right) +C }$$