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

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

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

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{1}{3 u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{4}

= \frac{1}{3} \cdot \frac{3}{4}

Simplify

= \frac{1}{4}

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