What is the integral from 0 to b of x/((2x^2+1)^2) ?

The integral from 0 to b of x/((2x^2+1)^2) is (b^2)/(2(2b^2+1))

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integral from 0 to b of x/((2x^2+1)^2)

\int_{0}^{b} \frac{x}{( 2 x ^{2} + 1 ) ^{2}} d x

\frac{b ^{2}}{2 ( 2 b ^{2} + 1 )}

Solution steps

\int_{0}^{b} \frac{x}{( 2 x ^{2} + 1 ) ^{2}} d x

Apply u-substitution

: \int_{1}^{2 b^{2} + 1} \frac{1}{4 u ^{2}} d u

= \int_{1}^{2 b^{2} + 1} \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{2 b^{2} + 1} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{2 b^{2} + 1}

= \frac{1}{4} [- \frac{1}{u}]_{1}^{2 b^{2} + 1}

Compute the boundaries:

- \frac{1}{2 b ^{2} + 1} + 1

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

Simplify

= \frac{b ^{2}}{2 ( 2 b ^{2} + 1 )}

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What is the integral from 0 to b of x/((2x^2+1)^2) ?
The integral from 0 to b of x/((2x^2+1)^2) is (b^2)/(2(2b^2+1))