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

The integral from 0 to 1 of x/(sqrt(3x+1)) is 8/27

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integral from 0 to 1 of x/(sqrt(3x+1))

\int_{0}^{1} \frac{x}{3 x + 1} d x

\frac{8}{27}

Decimal

0.29629 \dots

Solution steps

\int_{0}^{1} \frac{x}{3 x + 1} d x

Apply u-substitution

: \int_{1}^{2} \frac{2 ( u ^{2} - 1 )}{9} d u

= \int_{1}^{2} \frac{2 ( u ^{2} - 1 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int_{1}^{2} u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{9} (\int_{1}^{2} u^{2} d u - \int_{1}^{2} 1 d u)

\int_{1}^{2} u^{2} d u = \frac{7}{3}

\int_{1}^{2} 1 d u = 1

= \frac{2}{9} (\frac{7}{3} - 1)

Simplify

\frac{2}{9} (\frac{7}{3} - 1) : \frac{8}{27}

= \frac{8}{27}

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What is the integral from 0 to 1 of x/(sqrt(3x+1)) ?
The integral from 0 to 1 of x/(sqrt(3x+1)) is 8/27