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

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

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

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Apply u-substitution

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

= \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

= 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

= 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

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

Simplify

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

= \frac{8}{3}

\int_{0}^{π} ∣ sin (x) ∣ d x

\int_{1}^{e} \frac{2 x ^{2} + 1}{2 x} d x

\int_{0}^{2 π} cos (2 θ) d θ

\int_{0}^{\frac{π}{3}} sec (θ) tan (θ) d θ

\int_{0}^{2 π} 1 + x^{2} d x

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