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

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

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integral of ln(sqrt(x+1))

\int ln (x + 1) d x

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

Solution steps

\int ln (x + 1) d x

Simplify

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

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

Apply u-substitution

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

Apply Integration By Parts

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

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

Substitute back

u = x + 1

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

Expand

(x + 1) ln (x + 1) - (x + 1) : x ln (x + 1) + ln (x + 1) - x - 1

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

Add a constant to the solution

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

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