What is the integral of x/((1+4x^2)^2) ?

The integral of x/((1+4x^2)^2) is -1/(8(1+4x^2))+C

integral of x/((1+4x^2)^2)

\int \frac{x}{( 1 + 4 x ^{2} ) ^{2}} d x

- \frac{1}{8 ( 1 + 4 x ^{2} )} + C

Solution steps

\int \frac{x}{( 1 + 4 x ^{2} ) ^{2}} d x

Apply u-substitution

= \int \frac{1}{8 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= \frac{1}{8} (- \frac{1}{u})

Substitute back

u = 1 + 4 x^{2}

= \frac{1}{8} (- \frac{1}{1 + 4 x ^{2}})

Simplify

\frac{1}{8} (- \frac{1}{1 + 4 x ^{2}}) : - \frac{1}{8 ( 1 + 4 x ^{2} )}

= - \frac{1}{8 ( 1 + 4 x ^{2} )}

Add a constant to the solution

= - \frac{1}{8 ( 1 + 4 x ^{2} )} + C

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What is the integral of x/((1+4x^2)^2) ?
The integral of x/((1+4x^2)^2) is -1/(8(1+4x^2))+C