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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = ln (1 + x)

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

Add a constant to the solution

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

y^{^{'}} = y (y - 3), y (0) = a

n = 0 \sum \infty \frac{n x ^{n}}{n + 2}

f (x) = 5^{x^{2}}

\frac{\partial}{\partial y} (\frac{1}{x + y})

\frac{d y}{d x} = - \frac{( x + 6 )}{7 y ^{2}}

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