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

The integral from 1 to 4 of 1/((x+3)^2) is 3/28

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integral from 1 to 4 of 1/((x+3)^2)

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

\frac{3}{28}

Decimal Notation

0.10714 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{28}

= \frac{3}{28}

\int_{- 3}^{3} \frac{1}{x - 1} d x

\int_{2}^{3} (2 + 2 y)^{2} d y

\int_{0}^{1} - 1 + \frac{π x}{2} d x

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

\int_{3}^{4} - x ln (x - 3) d x

What is the integral from 1 to 4 of 1/((x+3)^2) ?
The integral from 1 to 4 of 1/((x+3)^2) is 3/28