What is the integral from 1 to e-1 of 2xln(x+1) ?

The integral from 1 to e-1 of 2xln(x+1) is 2((e^2)/4-1)

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integral from 1 to e-1 of 2xln(x+1)

\int_{1}^{e - 1} 2 x ln (x + 1) d x

2 (\frac{e ^{2}}{4} - 1)

Solution steps

\int_{1}^{e - 1} 2 x ln (x + 1) d x

Take the constant out:

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

= 2 \cdot \int_{1}^{e - 1} x ln (x + 1) d x

Apply u-substitution

: \int_{2}^{e} (u - 1) ln (u) d u

= 2 \cdot \int_{2}^{e} (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]_{2}^{e}

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{2}^{e}

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

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{e}

Simplify

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{e} : 2 [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{2}^{e}

= 2 [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{2}^{e}

Compute the boundaries:

\frac{e ^{2}}{4} - 1

= 2 (\frac{e ^{2}}{4} - 1)

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\int_{0}^{3} r d r

What is the integral from 1 to e-1 of 2xln(x+1) ?
The integral from 1 to e-1 of 2xln(x+1) is 2((e^2)/4-1)