What is the integral of x/((4x^2+1)^5) ?

The integral of x/((4x^2+1)^5) is -1/(32(4x^2+1)^4)+C

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integral of x/((4x^2+1)^5)

\int \frac{x}{( 4 x ^{2} + 1 ) ^{5}} d x

- \frac{1}{32 ( 4 x ^{2} + 1 ) ^{4}} + C

Solution steps

\int \frac{x}{( 4 x ^{2} + 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 ( 4 u + 1 ) ^{5}} d u

= \int \frac{1}{2 ( 4 u + 1 ) ^{5}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{( 4 u + 1 ) ^{5}} d u

Apply u-substitution

: \int \frac{1}{4 v ^{5}} d v

= \frac{1}{2} \cdot \int \frac{1}{4 v ^{5}} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{4 ( 4 x ^{2} + 1 ) ^{4}}) : - \frac{1}{32 ( 4 x ^{2} + 1 ) ^{4}}

= - \frac{1}{32 ( 4 x ^{2} + 1 ) ^{4}}

Add a constant to the solution

= - \frac{1}{32 ( 4 x ^{2} + 1 ) ^{4}} + C

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What is the integral of x/((4x^2+1)^5) ?
The integral of x/((4x^2+1)^5) is -1/(32(4x^2+1)^4)+C