Indefinite Integral - How to do questions with square roots?

Question

$$\int \frac{dx}{x^4 \sqrt{a^2 + x^2}}$$

In the above question, my first step would be to try and get out of the square root, so I would take $ t^2 = a^2 + x^2 $. But that gets me nowhere. How would you solve this, and if you are going to take a substitution what is the logic behind that substitution?

Answer 1

substitute $x=a\tan { \theta } ,dx=\frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } }$ so

$$\\ \\ \\ \int \frac { dx }{ x^{ 4 }\sqrt { a^{ 2 }+x^{ 2 } } } =\int \frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } { \left( a\tan { \theta } \right) }^{ 4 }\sqrt { a^{ 2 }+{ a }^{ 2 }\tan ^{ 2 }{ \theta } } } =\frac { 1 }{ { a }^{ 4 } } \int { \frac { \cos ^{ 3 }{ \theta } }{ \sin ^{ 4 }{ \theta } } } \, d\theta =\\ \\ =\frac { 1 }{ { a }^{ 4 } } \int { \frac { 1-\sin ^{ 2 }{ \theta } }{ \sin ^{ 4 }{ \theta } } d\sin { \theta } } =\frac { 1 }{ { a }^{ 4 } } \left[ \int { \frac { d\sin { \theta } }{ \sin ^{ 4 }{ \theta } } } -\int { \frac { d\sin { \theta } }{ \sin ^{ 2 }{ \theta } } } \right] =\\ =\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \theta } } +\frac { 1 }{ \sin { \theta } } \right] =\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \left( \arctan { \frac { x }{ a } } \right) } } +\frac { 1 }{ \sin { \left( \arctan { \frac { x }{ a } } \right) } } \right] +C\\ $$

Answer 2

Hint...try substituting $$x=a\sinh \theta$$ and using standard hyperbolic identities and derivatives

Answer 3

You can use trigonometric functions. Substitute $x=a*tan(u)$ and $dx=a*\frac{du}{{cos}^{2}u}$ this gives:

$$\int \dfrac{dx}{x^4 \sqrt{a^2 + x^2}} =\int \dfrac{du}{a^4\cos(u){\tan}^{4}(u)}$$

Working further on the solution of Battani and by simplifying:

$${ \sin{ \left( \arctan { \frac { x }{ a } } \right) } }=\frac{x}{\sqrt{x^2+a^2}}$$

This gives:

$$\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \left( \arctan { \frac { x }{ a } } \right) } } +\frac { 1 }{ \sin { \left( \arctan { \frac { x }{ a } } \right) } } \right] +C\\=\frac{1}{a^4}\left[ -\frac{(x^2+a^2)^{3/2}}{3x^3}+\frac{\sqrt{x^2+a^2}}{x}\right]$$

Simplifying further:

$$\frac{1}{a^4}\left[ -\frac{(3x^2-{x^2+a^2})\sqrt{x^2+a^2})}{3x^3}\right]=$$

$$-\frac{(2x^2+a^2)\sqrt{x^2+a^2})}{3a^4x^3}$$