What is the integral from 0 to 1 of (4x)/((x^2+1)^2) ?

The integral from 0 to 1 of (4x)/((x^2+1)^2) is 1

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integral from 0 to 1 of (4x)/((x^2+1)^2)

\int_{0}^{1} \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x

1

Solution steps

\int_{0}^{1} \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{2 u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

4 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{2} : 2 [- \frac{1}{u}]_{1}^{2}

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

Compute the boundaries:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

= 1

x \to 5 lim (\frac{3 x + 5}{5 x + 6})

\int \frac{3}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{in x}{x ^{2}} d x

\frac{d}{d t} (3 sin^{3} (t))

7 \frac{d y}{d x} = y^{2}

What is the integral from 0 to 1 of (4x)/((x^2+1)^2) ?
The integral from 0 to 1 of (4x)/((x^2+1)^2) is 1