What is the integral of x/((x^2+1)^{23/2)} ?

The integral of x/((x^2+1)^{23/2)} is -1/(21(x^2+1)^{21/2)}+C

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integral of x/((x^2+1)^{23/2)}

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{23}{2}}} d x

- \frac{1}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}} + C

Solution steps

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{23}{2}}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{23}{2}}} d u

= \int \frac{1}{2 u ^{\frac{23}{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{23}{2}}} d u

Apply the Power Rule

: - \frac{2}{21 u ^{\frac{21}{2}}}

= \frac{1}{2} (- \frac{2}{21 u ^{\frac{21}{2}}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (- \frac{2}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}}) : - \frac{1}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}}

= - \frac{1}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}}

Add a constant to the solution

= - \frac{1}{21 ( x ^{2} + 1 ) ^{\frac{21}{2}}} + C

x \to 10 - lim (ln (100 - x^{2}))

\frac{\partial}{\partial y} (2 y^{2})

x \to \infty + lim (i x)

\int_{0}^{\frac{π}{2}} 1

\frac{d}{d x} (\frac{2}{x + 8})

What is the integral of x/((x^2+1)^{23/2)} ?
The integral of x/((x^2+1)^{23/2)} is -1/(21(x^2+1)^{21/2)}+C