What is the integral from 0 to 1 of xsqrt(x+1) ?

The integral from 0 to 1 of xsqrt(x+1) is (4sqrt(2)+4)/(15)

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integral from 0 to 1 of xsqrt(x+1)

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

\frac{4 2 + 4}{15}

Decimal

0.64379 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= \int_{1}^{2} u^{\frac{3}{2}} - u 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

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

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

= \frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}

Simplify

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= \frac{4 2 + 4}{15}

\int_{0}^{x^{2}} 1 + t^{2} d t

\int_{- 1}^{1} 2 x + 3 d x

\int_{0}^{1} \frac{x ^{2}}{2} d x

\int_{- 3}^{1} \frac{1}{x ^{4}} d x

\int_{0}^{4} x x^{2} + 5 d x

What is the integral from 0 to 1 of xsqrt(x+1) ?
The integral from 0 to 1 of xsqrt(x+1) is (4sqrt(2)+4)/(15)