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

The integral of ln(1+x) is ln(1+x)+xln(1+x)-1-x+C

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integral of ln(1+x)

integral

ln (1 + x)

ln (1 + x) + x ln (1 + x) - 1 - x + C

Solution steps

\int ln (1 + x) 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 = 1 + x

= (1 + x) ln (1 + x) - (1 + x)

Simplify

(1 + x) ln (1 + x) - (1 + x) : ln (1 + x) + x ln (1 + x) - 1 - x

= ln (1 + x) + x ln (1 + x) - 1 - x

Add a constant to the solution

= ln (1 + x) + x ln (1 + x) - 1 - x + C

in t e g r a l 3 t + 4

in t e g r a l \frac{x ^{2}}{( x + 1 )}

in t e g r a l \frac{1}{( x - 2 )}

in t e g r a l \frac{t ^{2}}{( 1 - t ^{2} )}

in t e g r a l \frac{1}{( a - x )}

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