What is the integral from 0 to 2 of ln(9-x^2) ?

The integral from 0 to 2 of ln(9-x^2) is 5ln(5)-4

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integral from 0 to 2 of ln(9-x^2)

\int_{0}^{2} ln (9 - x^{2}) d x

5 ln (5) - 4

Decimal Notation

4.04718 \dots

Solution steps

\int_{0}^{2} ln (9 - x^{2}) d x

Apply Integration By Parts

: [x ln (9 - x^{2}) - \int - \frac{2 x ^{2}}{9 - x ^{2}} d x]_{0}^{2}

= [x ln (9 - x^{2}) - \int - \frac{2 x ^{2}}{9 - x ^{2}} d x]_{0}^{2}

\int - \frac{2 x ^{2}}{9 - x ^{2}} d x = - 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})

= [x ln (9 - x^{2}) - (- 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}))]_{0}^{2}

Simplify

= [x ln (9 - x^{2}) + 6 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})]_{0}^{2}

Compute the boundaries:

2 ln (5) + 6 (\frac{ln ( 3 ) + ln ( \frac{5}{3} )}{2} - \frac{2}{3})

= 2 ln (5) + 6 (\frac{ln ( 3 ) + ln ( \frac{5}{3} )}{2} - \frac{2}{3})

Simplify

= 5 ln (5) - 4

\int_{1}^{2} 4 x d x

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

\int_{0}^{1} e^{- s t} t^{2} d t

\int_{- 2}^{4} (2 x - \frac{x ^{2}}{2}) d x

\int_{- 1}^{2} (1 - t^{2}) t d t

What is the integral from 0 to 2 of ln(9-x^2) ?
The integral from 0 to 2 of ln(9-x^2) is 5ln(5)-4