What is the integral of x/((x^2+1)^{17/2)} ?

The integral of x/((x^2+1)^{17/2)} is -1/(15(x^2+1)^{15/2)}+C

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integral of x/((x^2+1)^{17/2)}

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{17}{2}}} d x

- \frac{1}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}} + C

Solution steps

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{17}{2}}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{17}{2}}} d u

= \int \frac{1}{2 u ^{\frac{17}{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 ^{\frac{17}{2}}} d u

Apply the Power Rule

: - \frac{2}{15 u ^{\frac{15}{2}}}

= \frac{1}{2} (- \frac{2}{15 u ^{\frac{15}{2}}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (- \frac{2}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}}) : - \frac{1}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}}

= - \frac{1}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}}

Add a constant to the solution

= - \frac{1}{15 ( x ^{2} + 1 ) ^{\frac{15}{2}}} + C

y^{^{'}} + y = - 4 cos (x)

x \to \infty lim (1^{x})

(7 x^{6} + 4 x^{3})^{4}

\int \frac{sin ( x ) cos ( x )}{5 + sin ( x )} d x

y^{^{''}} + 2 y^{^{'}} = e^{t}, y (0) = 1, y^{^{'}} (0) = 2

What is the integral of x/((x^2+1)^{17/2)} ?
The integral of x/((x^2+1)^{17/2)} is -1/(15(x^2+1)^{15/2)}+C