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

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

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

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

3 ln (3) - 2

Decimal

1.29583 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

= [u ln (u) - u]_{1}^{3}

Compute the boundaries:

3 ln (3) - 2

= 3 ln (3) - 2

\int_{0}^{\frac{π}{2}} \frac{1}{cos ^{2} ( x )} d x

\int_{0}^{\frac{π}{4}} cos (x) d x

\int_{- 2}^{2} x (x^{2} - 7) d x

\int_{0}^{π} e^{x} cos^{2} (x) d x

\int_{1}^{e} \frac{3}{x} d x

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