What is the integral of x/((x^2+1)^{7/2)} ?

The integral of x/((x^2+1)^{7/2)} is -1/(5(x^2+1)^{5/2)}+C

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integral of x/((x^2+1)^{7/2)}

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

- \frac{1}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( u + 1 ) ^{\frac{7}{2}}} d u

= \int \frac{1}{2 ( u + 1 ) ^{\frac{7}{2}}} 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 + 1 ) ^{\frac{7}{2}}} d u

Apply u-substitution

: \int \frac{1}{v ^{\frac{7}{2}}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{7}{2}}} d v

Apply the Power Rule

: - \frac{2}{5 v ^{\frac{5}{2}}}

= \frac{1}{2} (- \frac{2}{5 v ^{\frac{5}{2}}})

Substitute back

= \frac{1}{2} (- \frac{2}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}}) : - \frac{1}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}}

= - \frac{1}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}}

Add a constant to the solution

= - \frac{1}{5 ( x ^{2} + 1 ) ^{\frac{5}{2}}} + C

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What is the integral of x/((x^2+1)^{7/2)} ?
The integral of x/((x^2+1)^{7/2)} is -1/(5(x^2+1)^{5/2)}+C