What is the integral from 0 to 2 of 2xln(1+x) ?

The integral from 0 to 2 of 2xln(1+x) is 3ln(3)

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integral from 0 to 2 of 2xln(1+x)

\int_{0}^{2} 2 x ln (1 + x) d x

3 ln (3)

Decimal Notation

3.29583 \dots

Solution steps

\int_{0}^{2} 2 x ln (1 + x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x ln (1 + x) d x

Apply u-substitution

: \int_{1}^{3} (u - 1) ln (u) d u

= 2 \cdot \int_{1}^{3} (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]_{1}^{3}

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{3}

\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)]_{1}^{3}

Simplify

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{3} : 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

= 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{3}

Compute the boundaries:

\frac{3}{2} ln (3)

= 2 \cdot \frac{3}{2} ln (3)

Simplify

= 3 ln (3)

\int_{1}^{2} \frac{x ^{2}}{2} d x

\int_{1}^{3} x^{2} + 2 x - 5 d x

\int_{- \infty}^{0} \frac{3}{9 + x ^{2}} d x

\int_{0}^{1} x y^{2} d x

\int_{2}^{3} (x^{2} + x + 1) d x

What is the integral from 0 to 2 of 2xln(1+x) ?
The integral from 0 to 2 of 2xln(1+x) is 3ln(3)