What is the integral of 1/(x^2(1+x^2)^{3/2)} ?

The integral of 1/(x^2(1+x^2)^{3/2)} is -((1+x^2)^{1/2})/x-x/((1+x^2)^{1/2)}+C

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integral of 1/(x^2(1+x^2)^{3/2)}

\int \frac{1}{x ^{2} ( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

- \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{x ^{2} ( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{tan ^{2} ( u ) sec ( u )} d u

= \int \frac{1}{tan ^{2} ( u ) sec ( u )} d u

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{2}} d v

= \int \frac{1 - v ^{2}}{v ^{2}} d v

Expand

\frac{1 - v ^{2}}{v ^{2}} : \frac{1}{v ^{2}} - 1

= \int \frac{1}{v ^{2}} - 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{v ^{2}} d v - \int 1 d v

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int 1 d v = v

= - \frac{1}{v} - v

Substitute back

= - \frac{1}{sin ( arctan ( x ) )} - sin (arctan (x))

Simplify

- \frac{1}{sin ( arctan ( x ) )} - sin (arctan (x)) : - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} - \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

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What is the integral of 1/(x^2(1+x^2)^{3/2)} ?
The integral of 1/(x^2(1+x^2)^{3/2)} is -((1+x^2)^{1/2})/x-x/((1+x^2)^{1/2)}+C