What is the integral from 0 to 15 of x/(sqrt(x+1)) ?

The integral from 0 to 15 of x/(sqrt(x+1)) is 36

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integral from 0 to 15 of x/(sqrt(x+1))

\int_{0}^{15} \frac{x}{x + 1} d x

36

Solution steps

\int_{0}^{15} \frac{x}{x + 1} d x

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \int_{1}^{4} 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

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

\int_{1}^{4} u^{2} d u = 21

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

= 2 (21 - 3)

Simplify

= 36

- x^{3} + 3, y = x, x = - 1, x = 1

\int (x cos (x) + sin (x)) 1 + u d x

\int x^{3} - 4 x + 2 d x

\frac{d r}{d θ} = - π sin (π θ), r (2) = 3

\frac{d}{d x} (lo g_{7} (3 x^{2} + 1))

What is the integral from 0 to 15 of x/(sqrt(x+1)) ?
The integral from 0 to 15 of x/(sqrt(x+1)) is 36