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

The integral of 1/(x^2+sqrt(x)) is

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integral of 1/(x^2+sqrt(x))

\int \frac{1}{x ^{2} + x} d x

2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))) + C

Solution steps

\int \frac{1}{x ^{2} + x} d x

Apply u-substitution

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

= \int \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 \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 \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 \frac{1}{3 ( u + 1 )} d u + \int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u)

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3}))

= 2 (\frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3})))

Substitute back

u = x

= 2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 ( x ) ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

Simplify

2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 ( x ) ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))) : 2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

= 2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

Add a constant to the solution

= 2 (\frac{1}{3} ln x + 1 + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))) + C

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