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

: \int_{3}^{12} \frac{1}{u ^{2}} d u

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{4}

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

Simplify

= 1

\int_{0}^{1} x^{12} + 1 2^{x} d x

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

\int_{3}^{4} \frac{1}{x ^{2} - 3 x + 2} d x

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

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

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