What is the integral of x/((x^2+4)^{15)} ?

The integral of x/((x^2+4)^{15)} is -1/(28(x^2+4)^{14)}+C

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integral of x/((x^2+4)^{15)}

\int \frac{x}{( x ^{2} + 4 ) ^{15}} d x

- \frac{1}{28 ( x ^{2} + 4 ) ^{14}} + C

Solution steps

\int \frac{x}{( x ^{2} + 4 ) ^{15}} d x

Apply u-substitution

: \int \frac{1}{2 ( u + 4 ) ^{15}} d u

= \int \frac{1}{2 ( u + 4 ) ^{15}} 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}{( u + 4 ) ^{15}} d u

Apply u-substitution

: \int \frac{1}{v ^{15}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{15}} d v

Apply the Power Rule

: - \frac{1}{14 v ^{14}}

= \frac{1}{2} (- \frac{1}{14 v ^{14}})

Substitute back

= \frac{1}{2} (- \frac{1}{14 ( x ^{2} + 4 ) ^{14}})

Simplify

\frac{1}{2} (- \frac{1}{14 ( x ^{2} + 4 ) ^{14}}) : - \frac{1}{28 ( x ^{2} + 4 ) ^{14}}

= - \frac{1}{28 ( x ^{2} + 4 ) ^{14}}

Add a constant to the solution

= - \frac{1}{28 ( x ^{2} + 4 ) ^{14}} + C

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What is the integral of x/((x^2+4)^{15)} ?
The integral of x/((x^2+4)^{15)} is -1/(28(x^2+4)^{14)}+C