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

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

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

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

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

Decimal

1.67129 \dots

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{1 + u ^{3}} : \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})

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