What is the integral of x*ln(x-1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x*ln(x-1)

\int x \cdot ln (x - 1) d x

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

Solution steps

\int x ln (x - 1) d x

Apply u-substitution

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

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

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} + u) - \int \frac{1}{2} (u + 2) d u

= ln (u) (\frac{u ^{2}}{2} + u) - \int \frac{1}{2} (u + 2) d u

\int \frac{1}{2} (u + 2) d u = \frac{u ^{2}}{4} + u

= ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

2 x \frac{d y}{d x} - y = 3 x^{2}

x \to 2 lim (\frac{1}{2 x})

\int 3 x^{2} e^{9 x^{3}} d x

f (x) = \frac{x - 1}{x + 3}

\int_{0}^{\infty} e^{(- 2 x)} d x

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