What is the integral of xln(y+1) ?

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

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integral of xln(y+1)

\int x ln (y + 1) d y

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

Solution steps

\int x ln (y + 1) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= x \cdot \int ln (1 + y) d y

Apply u-substitution

: \int ln (u) d u

= x \cdot \int ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= x (u ln (u) - u)

Substitute back

u = 1 + y

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

Expand

(1 + y) ln (1 + y) - (1 + y) : ln (1 + y) + y ln (1 + y) - 1 - y

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

Add a constant to the solution

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

n = 0 \sum \infty (\frac{A}{A + 1})^{n}

f (x) = e^{9 x}

f (x) = 3 x^{2} + 4 x, at x = - 2

x \to 0 lim (cos (\frac{7}{x}))

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