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

The integral from 0 to 1 of x/(sqrt(1+3x)+1) is 5/27

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

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

\frac{5}{27}

Decimal

0.18518 \dots

Solution steps

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

Multiply by the conjugate of

1 + 3 x + 1 : \frac{3 x}{1 + 3 x - 1}

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

Take the constant out:

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

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

Simplify

\frac{x}{\frac{x}{1 + 3 x - 1}} : 3 x + 1 - 1

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

Apply the Sum Rule:

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

= \frac{1}{3} (\int_{0}^{1} 3 x + 1 d x - \int_{0}^{1} 1 d x)

\int_{0}^{1} 3 x + 1 d x = \frac{14}{9}

\int_{0}^{1} 1 d x = 1

= \frac{1}{3} (\frac{14}{9} - 1)

Simplify

\frac{1}{3} (\frac{14}{9} - 1) : \frac{5}{27}

= \frac{5}{27}

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What is the integral from 0 to 1 of x/(sqrt(1+3x)+1) ?
The integral from 0 to 1 of x/(sqrt(1+3x)+1) is 5/27