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

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

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

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

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

Decimal

0.35208 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} (u - 2) ln (u) d u

= \int_{1}^{3} (u - 2) ln (u) d u

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

\int_{1.7}^{4.7} 2 t^{\frac{1}{2}} d t

\int_{- 2}^{2} 2^{x} d x

\int_{4}^{6} 700 x 3 + 2 x d x

\int_{0}^{l n (2)} e^{- 3 y} d y

\int_{0}^{4} \frac{1}{4 t ^{2} + 81} d t

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