What is the integral of (x^4)/((1+x^2)^{3/2)} ?

The integral of (x^4)/((1+x^2)^{3/2)} is

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integral of (x^4)/((1+x^2)^{3/2)}

\int \frac{x ^{4}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + C

Solution steps

\int \frac{x ^{4}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int tan^{4} (u) cos (u) d u

= \int tan^{4} (u) cos (u) d u

Rewrite using trig identities

= \int \frac{sin ^{4} ( u )}{cos ^{3} ( u )} d u

Rewrite using trig identities

= \int tan^{3} (u) sin (u) d u

Apply u-substitution

: \int \frac{v ^{4}}{( 1 - v ^{2} ) ^{2}} d v

= \int \frac{v ^{4}}{( 1 - v ^{2} ) ^{2}} d v

Take the partial fraction of

\frac{v ^{4}}{( 1 - v ^{2} ) ^{2}} : 1 - \frac{3}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{3}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}}

= \int 1 - \frac{3}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{3}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}} 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 1 d v - \int \frac{3}{4 ( v + 1 )} d v + \int \frac{1}{4 ( v + 1 ) ^{2}} d v + \int \frac{3}{4 ( v - 1 )} d v + \int \frac{1}{4 ( v - 1 ) ^{2}} d v

\int 1 d v = v

\int \frac{3}{4 ( v + 1 )} d v = \frac{3}{4} ln ∣ v + 1 ∣

\int \frac{1}{4 ( v + 1 ) ^{2}} d v = - \frac{1}{4 ( v + 1 )}

\int \frac{3}{4 ( v - 1 )} d v = \frac{3}{4} ln ∣ v - 1 ∣

\int \frac{1}{4 ( v - 1 ) ^{2}} d v = - \frac{1}{4 ( v - 1 )}

= v - \frac{3}{4} ln ∣ v + 1 ∣ - \frac{1}{4 ( v + 1 )} + \frac{3}{4} ln ∣ v - 1 ∣ - \frac{1}{4 ( v - 1 )}

Substitute back

= sin (arctan (x)) - \frac{3}{4} ln ∣ sin (arctan (x)) + 1 ∣ - \frac{1}{4 ( sin ( arctan ( x ) ) + 1 )} + \frac{3}{4} ln ∣ sin (arctan (x)) - 1 ∣ - \frac{1}{4 ( sin ( arctan ( x ) ) - 1 )}

Simplify

sin (arctan (x)) - \frac{3}{4} ln ∣ sin (arctan (x)) + 1 ∣ - \frac{1}{4 ( sin ( arctan ( x ) ) + 1 )} + \frac{3}{4} ln ∣ sin (arctan (x)) - 1 ∣ - \frac{1}{4 ( sin ( arctan ( x ) ) - 1 )} : \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )}

= \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )}

Add a constant to the solution

= \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x + ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + \frac{3}{4} ln \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} - 1 - \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{4 ( x - ( 1 + x ^{2} ) ^{\frac{1}{2}} )} + C

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What is the integral of (x^4)/((1+x^2)^{3/2)} ?
The integral of (x^4)/((1+x^2)^{3/2)} is