What is the integral of 1/(x^4sqrt(x^2+1)) ?

The integral of 1/(x^4sqrt(x^2+1)) is -((1+x^2)^{3/2}}{3x^3}+\frac{sqrt(1+x^2))/x+C

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integral of 1/(x^4sqrt(x^2+1))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cot^{4} (u) sec (u) d u

= \int cot^{4} (u) sec (u) d u

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

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

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

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

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

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

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

= - \frac{1}{3 v ^{3}} - (- \frac{1}{v})

Substitute back

= - \frac{1}{3 sin ^{3} ( arctan ( x ) )} - (- \frac{1}{sin ( arctan ( x ) )})

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(x^4sqrt(x^2+1)) ?
The integral of 1/(x^4sqrt(x^2+1)) is -((1+x^2)^{3/2}}{3x^3}+\frac{sqrt(1+x^2))/x+C