What is the integral from 0 to 1 of 1/(x^2+sqrt(x)) ?

The integral from 0 to 1 of 1/(x^2+sqrt(x)) is 2(1/3 ln(2)+pi/(3sqrt(3)))

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integral from 0 to 1 of 1/(x^2+sqrt(x))

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

2 (\frac{1}{3} ln (2) + \frac{π}{3 3})

Decimal Notation

1.67129 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{2}{u ^{3} + 1} d u

= \int_{0}^{1} \frac{2}{u ^{3} + 1} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{u ^{3} + 1} d u

Take the partial fraction of

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

= 2 \cdot \int_{0}^{1} \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 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_{0}^{1} \frac{1}{3 ( u + 1 )} d u + \int_{0}^{1} \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u)

\int_{0}^{1} \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln (2)

\int_{0}^{1} \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{π}{3 3}

= 2 (\frac{1}{3} ln (2) + \frac{π}{3 3})

\int_{0}^{2} \frac{2 v}{v + 1} d v

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\int_{a}^{b} x 2 x - x^{2} d x

What is the integral from 0 to 1 of 1/(x^2+sqrt(x)) ?
The integral from 0 to 1 of 1/(x^2+sqrt(x)) is 2(1/3 ln(2)+pi/(3sqrt(3)))