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

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

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integral of (ln(sqrt(x)+1))/(x+sqrt(x))

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

ln^{2} (1 + x) + C

Solution steps

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

Apply u-substitution

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

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

\frac{ln ( u + 1 )}{u + 1} = ln (u + 1) \frac{1}{u + 1}

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

Apply u-substitution

: \int v d v

= 2 \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= 2 \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

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

= ln^{2} (1 + x)

Add a constant to the solution

= ln^{2} (1 + x) + C

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