What is the integral from 6 to 12 of y/(y^2-2y-3) ?

The integral from 6 to 12 of y/(y^2-2y-3) is (ln(13)-ln(7)+3ln(3))/4

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integral from 6 to 12 of y/(y^2-2y-3)

\int_{6}^{12} \frac{y}{y ^{2} - 2 y - 3} d y

\frac{ln ( 13 ) - ln ( 7 ) + 3 ln ( 3 )}{4}

Decimal Notation

0.97871 \dots

Solution steps

\int_{6}^{12} \frac{y}{y ^{2} - 2 y - 3} d y

Take the partial fraction of

\frac{y}{y ^{2} - 2 y - 3} : \frac{1}{4 ( y + 1 )} + \frac{3}{4 ( y - 3 )}

= \int_{6}^{12} \frac{1}{4 ( y + 1 )} + \frac{3}{4 ( y - 3 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{6}^{12} \frac{1}{4 ( y + 1 )} d y + \int_{6}^{12} \frac{3}{4 ( y - 3 )} d y

\int_{6}^{12} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (ln (13) - ln (7))

\int_{6}^{12} \frac{3}{4 ( y - 3 )} d y = \frac{3}{4} ln (3)

= \frac{1}{4} (ln (13) - ln (7)) + \frac{3}{4} ln (3)

Simplify

\frac{1}{4} (ln (13) - ln (7)) + \frac{3}{4} ln (3) : \frac{ln ( 13 ) - ln ( 7 ) + 3 ln ( 3 )}{4}

= \frac{ln ( 13 ) - ln ( 7 ) + 3 ln ( 3 )}{4}

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What is the integral from 6 to 12 of y/(y^2-2y-3) ?
The integral from 6 to 12 of y/(y^2-2y-3) is (ln(13)-ln(7)+3ln(3))/4