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

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

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integral of ln(1+sqrt(x))

\int ln (1 + x) d x

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

Solution steps

\int ln (1 + x) d x

Apply u-substitution

: \int ln (u) \cdot 2 (u - 1) d u

= \int ln (u) \cdot 2 (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 ln (u) (u - 1) d u

Apply Integration By Parts

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

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

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