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How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$ My work:

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

integral becomes $\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$

$\int \dfrac{\tan\theta(\tan^2\theta+1)+\tan \theta-7}{\sec\theta}\sec^2\theta d\theta$

$\int (\tan\theta(\sec^2\theta)+\tan \theta-7)\sec\theta d\theta$

$\int \tan\theta\sec^3\theta\ d\theta+\int \sec\theta \tan \theta\ d\theta-7\int \sec\theta d\theta$

$\int \tan\theta\sec^3\theta+\sec\theta -7\ln|\sec\theta+\tan\theta|+C$

I got stuck here in solving first part of above integral. I can't see the way to solve it. please help me solve it by substitution or other method. thanks

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    $\begingroup$ Excuse me what is the integral? The first of the title or $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx$? $\endgroup$
    – Sebastiano
    Jul 6, 2020 at 23:25
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    $\begingroup$ There's a 1 at the end of the title and a 7 at the end of the problem statement. That's the discrepancy @Sebastiano is pointing out. $\endgroup$ Jul 7, 2020 at 0:07
  • $\begingroup$ @RobertShore: that was my mistake. thank you sir. $\endgroup$
    – user805532
    Jul 7, 2020 at 0:46
  • $\begingroup$ @RobertShore I have deleted my answer considering that there is $-7$ instead of $-1$ even if the resolution criterion is the same. $\endgroup$
    – Sebastiano
    Jul 7, 2020 at 11:17

3 Answers 3

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

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

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    $\begingroup$ I like your cleverness in solving the exercises.😉😉😉😉😉 $\endgroup$
    – Sebastiano
    Jul 7, 2020 at 11:18
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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

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

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  • $\begingroup$ can you give final answer if you substitute back to $x$? $\endgroup$
    – user805532
    Jul 7, 2020 at 1:00
  • $\begingroup$ Yes. I've edited to add. $\endgroup$ Jul 7, 2020 at 4:31
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\begin{aligned} \frac{x^{3}+2 x-7}{\sqrt{x^{2}+1}} d x &=\int \frac{x^{3}+2 x-7}{x} d \sqrt{x^{2}+1} \\ &=\int\left(x^{2}+2-\frac{7}{x}\right) d \sqrt{x^{2}+1} \\ &=\int\left(x^{2}+1\right) d \sqrt{x^{2}+1}+\sqrt{x^{2}+1}-7 \int \frac{d \sqrt{x^{2}+1}}{x}\\ &=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}-7 \int \frac{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}+\frac{7}{2} \ln \left|\frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}-1}\right| +C\\ &=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}+4\right)+\frac{7}{2} \ln \left|\frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}-1}\right|+C \end{aligned}

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