What is the integral from 0 to 9 of 4/((3+x)^2) ?

The integral from 0 to 9 of 4/((3+x)^2) is 1

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integral from 0 to 9 of 4/((3+x)^2)

\int_{0}^{9} \frac{4}{( 3 + x ) ^{2}} d x

1

Solution steps

\int_{0}^{9} \frac{4}{( 3 + x ) ^{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}^{9} \frac{1}{( 3 + x ) ^{2}} d x

Apply u-substitution

= 4 \cdot \int_{3}^{12} \frac{1}{u ^{2}} d u

Apply the Power Rule

= 4 [- \frac{1}{u}]_{3}^{12}

Compute the boundaries:

\frac{1}{4}

= 4 \cdot \frac{1}{4}

Simplify

= 1

What is the integral from 0 to 9 of 4/((3+x)^2) ?
The integral from 0 to 9 of 4/((3+x)^2) is 1