What is the integral from 1 to 2 of (2x)/((x^2+3)^2) ?

The integral from 1 to 2 of (2x)/((x^2+3)^2) is 3/28

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integral from 1 to 2 of (2x)/((x^2+3)^2)

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

\frac{3}{28}

Decimal

0.10714 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{1}^{2} \frac{x}{( x ^{2} + 3 ) ^{2}} d x

Apply u-substitution

: \int_{4}^{7} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{4}^{7} \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

= 2 \cdot \frac{1}{2} \cdot \int_{4}^{7} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{4}^{7}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{7}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{7} : [- \frac{1}{u}]_{4}^{7}

= [- \frac{1}{u}]_{4}^{7}

Compute the boundaries:

\frac{3}{28}

= \frac{3}{28}

\int_{- 1}^{3} x^{2} - 2 x + 3 d x

\int_{2}^{5} (x + 5) d x

\int_{6}^{7} e^{6 - x} d x

\int_{0}^{3} 3^{2} - x^{2} d x

\int_{0}^{1} x^{2} + 2 d x

What is the integral from 1 to 2 of (2x)/((x^2+3)^2) ?
The integral from 1 to 2 of (2x)/((x^2+3)^2) is 3/28