How can I evaluate $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\mathrm dx?$

Question

How can I evaluate the following integral $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\mathrm dx$$ My work:

I substituted $x=\tan\theta$, $\mathrm dx=\sec^2\theta \mathrm d\theta $

The integral becomes $$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta\mathrm d\theta$$ $$=\int \dfrac{\tan\theta(\tan^2\theta+1)+\tan \theta-7}{\sec\theta}\sec^2\theta \mathrm d\theta$$ $$=\int (\tan\theta(\sec^2\theta)+\tan \theta-7)\sec\theta \mathrm d\theta$$ $$=\int \tan\theta\sec^3\theta\mathrm d\theta+\int \sec\theta \tan \theta\mathrm d\theta-7\int \sec\theta \mathrm d\theta$$ $$=\int \tan\theta\sec^3\theta\mathrm d\theta+\sec\theta -7\ln|\sec\theta+\tan\theta|+C$$

I got stuck here in evaluating the above integral. I can't see any way to evaluate it. Please help me evaluate it by substitution or some other method.

Answer 1

$$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx$$ $$=\int \frac{x(x^2+1)+x-7}{\sqrt{x^2+1}}\ dx$$ $$=\int x\sqrt{x^2+1}\ dx+\int \frac{x}{\sqrt{x^2+1}}\ dx-\int \frac{7}{\sqrt{x^2+1}}\ dx$$

$$=\frac12\int \sqrt{x^2+1}\ d(x^2+1)+\frac12\int \frac{d(x^2+1)}{\sqrt{x^2+1}}-7\int \frac{dx}{\sqrt{x^2+1}}$$ $$=\frac13(x^2+1)^{3/2}+\sqrt{x^2+1}-7\sinh^{-1}(x)+C$$

Answer 2

You're almost there.

$$\int \tan \theta \sec^3 \theta ~d \theta = \int \frac {\sin \theta}{\cos^4 \theta} ~d \theta = -\int \frac{du}{u^4}=\frac 13 u^{-3} = \frac {\sec^3 \theta}{3}+ C$$

Since $x= \tan \theta, x^2+1=\sec^2 \theta$, so

$$\frac{\sec^3 \theta}{3} + C = \frac{\sqrt{(x^2+1)^3}}{3}+C.$$

Answer 3

Perhaps the fastest way to evaluate the integral that you found problematic is:

$$\begin{align}\int\tan\theta\sec^3\theta\,\mathrm d\theta&=\int\sec^2\theta(\sec\theta\tan\theta\,\mathrm d\theta)\\&=\int\sec^2\theta\,\mathrm d(\sec\theta)\\&=\frac{\sec^3\theta}3+C\end{align}$$

Answer 4

\begin{aligned} \int\frac{x^{3}+2 x-7}{\sqrt{x^{2}+1}}\mathrm d x &=\int \frac{x^{3}+2 x-7}{x}\mathrm d\sqrt{x^{2}+1} \\ &=\int\left(x^{2}+2-\frac{7}{x}\right)\mathrm d \sqrt{x^{2}+1} \\ &=\int\left(x^{2}+1\right)\mathrm d \sqrt{x^{2}+1}+\sqrt{x^{2}+1}-7 \int \frac{\mathrm d \sqrt{x^{2}+1}}{x}\\ &=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}-7 \int \frac{\mathrm d \sqrt{x^{2}+1}}{\sqrt{\left(\sqrt{x^{2}+1}\right)^{2}-1}} \\ &=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}+7\ln\left|\sqrt{x^2+1}+\sqrt{\left(\sqrt{x^2+1}\right)^2-1}\right| +C\\ &=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}+4\right)+7\ln\left(x+\sqrt{x^2+1}\right)+C \end{aligned}

Answer 5

An alternative approach is to substitute $x=\sinh t$ since the derivative of $\sinh^{-1}x$ is present in the integrand:

$$\begin{align}\int\frac{x^3+2x-7}{\sqrt{x^2+1}}\mathrm dx&=\int\frac{\sinh3t-3\sinh t}4+2\sinh t-7\mathrm dt\\&=\frac{4\cosh^3t-3\cosh t}{12}+\frac54\cosh t-7t+C\\&=\frac{(x^2+1)^\frac32}3+\sqrt{x^2+1}-7\sinh^{-1}x+C\end{align}$$

Answer 6

The given integral is $$\int\frac{x^{3}+2x-7}{\sqrt{x^{2}+1}}dx$$

Let's assume that $$I=\int\frac{x^{3}+2x-7}{\sqrt{x^{2}+1}}dx$$

$$\implies I=\int\frac{x^{3}}{\sqrt{x^{2}+1}}dx+\int\frac{2x}{\sqrt{x^{2}+1}}dx-7\int\frac{1}{\sqrt{x^{2}+1}}dx$$

Just I seperated the numerator. Hope you will understand it.

Now let's assume that $$I_{1}=\int\frac{x^{3}}{\sqrt{x^{2}+1}}dx$$

$$I_{2}=\int\frac{2x}{\sqrt{x^{2}+1}}dx$$

$$I_{3}=\int\frac{1}{\sqrt{x^{2}+1}}dx$$

That is $$I=I_{1}+I_{2}+I_{3}$$

Now

$I_{1}=\int\frac{x^{3}}{\sqrt{x^{2}+1}}dx$

$\implies I_{1}=\int\frac{x^{2}x}{\sqrt{1+x^{2}}}dx$

Now let's substitute $(x^{2}+1)=u$

$\implies 2xdx=du$

$\implies I_{1}=\int\frac{u-1}{2\sqrt{u}}du$

Hope $I_{1}$ can be easily integrated now.

Again

$I_{2}=\int\frac{2x}{\sqrt{x^{2}+1}}dx$

$\implies I_{2}=2\int\frac{x}{\sqrt{x^{2}+1}}dx$

Now again substitute $(x^{2}+1)=t$ which implies $2xdx=dt$.

$\implies I_{2}=\int\frac{1}{\sqrt{t}}dt$

Now evaluate $I_{2}$. Hope it will be done

Again

$I_{3}=\int\frac{-7}{\sqrt{x^{2}+1}}dx$

For this I will do the famous substitution $x=\tan(z)$

Which implies $dx=\sec^{2}(z)dz$

$\implies I_{3}=-7\int\frac{\sec^{2}(z)}{\sec(z)}dz$

$\implies I_{3}=-7\int\sec(z)dz$

Because $\sqrt{1+\tan^{2}(z)}=|\sec(z)|=sgn(\sec(z))×\sec(z)$

$$\int\sec(z)dz=\ln|\sec(z)+\tan(z)|+C$$

Finally just add $I_{1},I_{2},I_{3}$ to get $I$.