What is the integral of 1/(x^4sqrt(16+x^2)) ?

The integral of 1/(x^4sqrt(16+x^2)) is ((x^2-8)sqrt(x^2+16))/(384x^3)+C

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integral of 1/(x^4sqrt(16+x^2))

\int \frac{1}{x ^{4} 16 + x ^{2}} d x

\frac{( x ^{2} - 8 ) x ^{2} + 16}{384 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} 16 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{256 tan ^{4} ( u )} d u

= \int \frac{sec ( u )}{256 tan ^{4} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{256} \cdot \int \frac{sec ( u )}{tan ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{256} \cdot \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

= \frac{1}{256} \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

Expand

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

= \frac{1}{256} \cdot \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

= \frac{1}{256} (\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}{256} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= \frac{1}{256} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{4} x ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}))

Simplify

\frac{1}{256} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{4} x ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})) : \frac{( x ^{2} - 8 ) x ^{2} + 16}{384 x ^{3}}

= \frac{( x ^{2} - 8 ) x ^{2} + 16}{384 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} - 8 ) x ^{2} + 16}{384 x ^{3}} + C

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

\int \frac{x}{1 - x} d x

\int (e^{3 x} + \frac{1}{x ^{3}} - 7 x^{\frac{3}{2}}) d x

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

What is the integral of 1/(x^4sqrt(16+x^2)) ?
The integral of 1/(x^4sqrt(16+x^2)) is ((x^2-8)sqrt(x^2+16))/(384x^3)+C