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

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

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integral of ln(sqrt(x+x^2))

\int ln (x + x^{2}) d x

\frac{1}{2} x ln (x + x^{2}) - x + \frac{1}{2} ln ∣ x + 1 ∣ + C

Solution steps

\int ln (x + x^{2}) d x

Simplify

ln (x + x^{2}) : \frac{1}{2} ln (x + x^{2})

= \int \frac{1}{2} ln (x + x^{2}) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int ln (x + x^{2}) d x

Apply Integration By Parts

: x ln (x + x^{2}) - \int \frac{1 + 2 x}{1 + x} d x

= \frac{1}{2} (x ln (x + x^{2}) - \int \frac{1 + 2 x}{1 + x} d x)

\int \frac{1 + 2 x}{1 + x} d x = 2 x - ln ∣ x + 1 ∣

= \frac{1}{2} (x ln (x + x^{2}) - (2 x - ln ∣ x + 1 ∣))

Simplify

\frac{1}{2} (x ln (x + x^{2}) - (2 x - ln ∣ x + 1 ∣)) : \frac{1}{2} x ln (x + x^{2}) - x + \frac{1}{2} ln ∣ x + 1 ∣

= \frac{1}{2} x ln (x + x^{2}) - x + \frac{1}{2} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{2} x ln (x + x^{2}) - x + \frac{1}{2} ln ∣ x + 1 ∣ + C

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