What is the integral of ln(x+a) ?

The integral of ln(x+a) is xln(x+a)+aln(x+a)-x-a+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+a)

\int ln (x + a) d x

x ln (x + a) + a ln (x + a) - x - a + C

Solution steps

\int ln (x + a) d x

Apply u-substitution

: \int ln (u) d u

= \int ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= u ln (u) - u

Substitute back

u = x + a

= (x + a) ln (x + a) - (x + a)

Simplify

(x + a) ln (x + a) - (x + a) : x ln (x + a) + a ln (x + a) - x - a

= x ln (x + a) + a ln (x + a) - x - a

Add a constant to the solution

= x ln (x + a) + a ln (x + a) - x - a + C

\int 5 + sec^{2} (x) d x

\int sin^{3} (\frac{x}{2}) cos^{7} (\frac{x}{2}) d x

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

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

\int \frac{( 1 + tan ( x ) ) ^{2}}{cos ^{2} ( x )} d x

What is the integral of ln(x+a) ?
The integral of ln(x+a) is xln(x+a)+aln(x+a)-x-a+C