What is the integral from 0 to 1 of (5r)/((4+r^2)^2) ?

The integral from 0 to 1 of (5r)/((4+r^2)^2) is 1/8

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integral from 0 to 1 of (5r)/((4+r^2)^2)

\int_{0}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\frac{1}{8}

Decimal Notation

0.125

Solution steps

\int_{0}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{r}{( 4 + r ^{2} ) ^{2}} d r

Apply u-substitution

: \int_{4}^{5} \frac{1}{2 u ^{2}} d u

= 5 \cdot \int_{4}^{5} \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

= 5 \cdot \frac{1}{2} \cdot \int_{4}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{4}^{5}

= 5 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5}

Simplify

5 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5} : \frac{5}{2} [- \frac{1}{u}]_{4}^{5}

= \frac{5}{2} [- \frac{1}{u}]_{4}^{5}

Compute the boundaries:

\frac{1}{20}

= \frac{5}{2} \cdot \frac{1}{20}

Simplify

= \frac{1}{8}

\int_{1}^{1} 1

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\int_{0}^{4} (x^{2} - 2 x) d x

What is the integral from 0 to 1 of (5r)/((4+r^2)^2) ?
The integral from 0 to 1 of (5r)/((4+r^2)^2) is 1/8