What is the integral of (ln(x+1))/(2(x+1)) ?

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

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integral of (ln(x+1))/(2(x+1))

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

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

Solution steps

\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

: \int u d u

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

Apply the Power Rule

: \frac{u ^{2}}{2}

= \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

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\frac{d}{d x} (2 e^{- 2})

\frac{\partial}{\partial t} (r t)

f (x) = - x^{2} + 4 x + 2, g (x) = x + 2

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