What is the integral of 1/(x+xsqrt(x)) ?

The integral of 1/(x+xsqrt(x)) is 2(ln|sqrt(x)|-ln|sqrt(x)+1|)+C

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integral of 1/(x+xsqrt(x))

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

2 (ln x - ln x + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ( u + 1 )} d u

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

Take the partial fraction of

\frac{1}{u ( u + 1 )} : \frac{1}{u} - \frac{1}{u + 1}

= 2 \cdot \int \frac{1}{u} - \frac{1}{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}{u} d u - \int \frac{1}{u + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 (ln ∣ u ∣ - ln ∣ u + 1 ∣)

Substitute back

u = x

= 2 (ln x - ln x + 1)

Add a constant to the solution

= 2 (ln x - ln x + 1) + C

\int 8 x + \frac{1}{x} d x

\frac{d}{d x} (\frac{x + 4}{3 x ^{2} e ^{x}})

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

\frac{d}{d x} (x + a)

\frac{d}{d x} (ln (x - y))

What is the integral of 1/(x+xsqrt(x)) ?
The integral of 1/(x+xsqrt(x)) is 2(ln|sqrt(x)|-ln|sqrt(x)+1|)+C