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

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

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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

Apply Integration By Parts

= 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

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