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

The integral of (ln(sqrt(x+1)))/(x+1) is 1/4 ln^2(1+x)+C

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

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

\frac{1}{4} ln^{2} (1 + x) + C

Solution steps

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

Simplify

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

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

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

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

Apply u-substitution

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (1 + x)

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( 1 + x )}{2} : \frac{1}{4} ln^{2} (1 + x)

= \frac{1}{4} ln^{2} (1 + x)

Add a constant to the solution

= \frac{1}{4} ln^{2} (1 + x) + C

What is the integral of (ln(sqrt(x+1)))/(x+1) ?
The integral of (ln(sqrt(x+1)))/(x+1) is 1/4 ln^2(1+x)+C