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

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

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integral of x/((1+x^2)^y)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{- y + 1}}{- y + 1}

= \frac{1}{2} \cdot \frac{u ^{- y + 1}}{- y + 1}

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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