Computing Indefinite Integrals — Question 5

Question 5

Evaluate $$\int \frac{x^2}{\sqrt{x^2+1}}\,dx.$$

Question 5 - Solution

Step 1: Rewrite the integrand

Write $x^2=(x^2+1)-1$: $$\frac{x^2}{\sqrt{x^2+1}} =\frac{(x^2+1)-1}{\sqrt{x^2+1}} =\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}.$$

Thus, $$\int \frac{x^2}{\sqrt{x^2+1}}\,dx =\int \sqrt{x^2+1}\,dx -\int \frac{1}{\sqrt{x^2+1}}\,dx.$$

Step 2: Trigonometric substitution

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

Step 3: Evaluate $\displaystyle \int \sqrt{x^2+1}\,dx$

Substitute: $$\int \sqrt{x^2+1}\,dx =\int \sec\theta\,(\sec^2\theta\,d\theta) =\int \sec^3\theta\,d\theta.$$

To evaluate $\int \sec^3\theta\,d\theta$, write $$\sec^3\theta=\sec\theta\sec^2\theta.$$ Let $$u=\sec\theta, \qquad dv=\sec^2\theta\,d\theta.$$ Then $$du=\sec\theta\tan\theta\,d\theta, \qquad v=\tan\theta.$$ By integration by parts, $$\int \sec^3\theta\,d\theta =\sec\theta\tan\theta-\int \tan\theta(\sec\theta\tan\theta)\,d\theta.$$ Simplify the integrand: $$\tan^2\theta=\sec^2\theta-1,$$ so $$\int \sec^3\theta\,d\theta =\sec\theta\tan\theta-\int \sec\theta(\sec^2\theta-1)\,d\theta.$$ Distribute: $$=\sec\theta\tan\theta-\int \sec^3\theta\,d\theta+\int \sec\theta\,d\theta.$$ Move the repeated integral to the left: $$2\int \sec^3\theta\,d\theta =\sec\theta\tan\theta+\int \sec\theta\,d\theta.$$

Now compute $\int \sec\theta\,d\theta$: $$\int \sec\theta\,d\theta =\int \frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}\,d\theta.$$ Let $$u=\sec\theta+\tan\theta, \qquad du=(\sec\theta\tan\theta+\sec^2\theta)\,d\theta =\sec\theta(\sec\theta+\tan\theta)\,d\theta.$$ Hence, $$\int \sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C.$$

Substitute back: $$2\int \sec^3\theta\,d\theta =\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|.$$ Therefore, $$\int \sec^3\theta\,d\theta =\frac12\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C.$$

Returning to $x$: $$\sec\theta=\sqrt{x^2+1}, \qquad \tan\theta=x.$$ Thus, $$\int \sqrt{x^2+1}\,dx =\frac12\left( x\sqrt{x^2+1} +\ln\!\left(x+\sqrt{x^2+1}\right) \right)+C.$$

Step 4: Evaluate $\displaystyle \int \frac{1}{\sqrt{x^2+1}}\,dx$

Substitute again: $$\int \frac{1}{\sqrt{x^2+1}}\,dx =\int \frac{1}{\sec\theta}(\sec^2\theta\,d\theta) =\int \sec\theta\,d\theta.$$ From above, $$\int \sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C =\ln\!\left(x+\sqrt{x^2+1}\right)+C.$$

Step 5: Combine results

$$\int \frac{x^2}{\sqrt{x^2+1}}\,dx = \frac12\left( x\sqrt{x^2+1} +\ln\!\left(x+\sqrt{x^2+1}\right) \right) - \ln\!\left(x+\sqrt{x^2+1}\right) + C.$$ Combine logarithms: $$=\frac{x}{2}\sqrt{x^2+1} -\frac12\ln\!\left(x+\sqrt{x^2+1}\right) + C.$$

Final Answer: $$\boxed{ \int \frac{x^2}{\sqrt{x^2+1}}\,dx = \frac{x}{2}\sqrt{x^2+1} -\frac12\ln\!\left(x+\sqrt{x^2+1}\right) + C }$$