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

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

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

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

8

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int_{1}^{2} 2 (u^{2} - 1) d u

Take the constant out:

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

= 3 \cdot 2 \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

= 3 \cdot 2 (\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

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

Simplify

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

= 8

\int_{0}^{x + y} e^{x} (y + 2 z) d z

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

\int_{0}^{1} x (x + y) d x

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

\int_{0}^{6} 5 d x

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