What is the integral from 0 to 3 of 8/((1+x)^2) ?

The integral from 0 to 3 of 8/((1+x)^2) is 6

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integral from 0 to 3 of 8/((1+x)^2)

\int_{0}^{3} \frac{8}{( 1 + x ) ^{2}} d x

6

Solution steps

\int_{0}^{3} \frac{8}{( 1 + x ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{3} \frac{1}{( 1 + x ) ^{2}} d x

Apply u-substitution

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

= 8 \cdot \int_{1}^{4} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

= 8 [- \frac{1}{u}]_{1}^{4}

Compute the boundaries:

\frac{3}{4}

= 8 \cdot \frac{3}{4}

Simplify

= 6

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What is the integral from 0 to 3 of 8/((1+x)^2) ?
The integral from 0 to 3 of 8/((1+x)^2) is 6