What is the integral from 7 to 12 of y/(y^2-3y-4) ?

The integral from 7 to 12 of y/(y^2-3y-4) is (ln(13)+9ln(2)-4ln(3))/5

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integral from 7 to 12 of y/(y^2-3y-4)

\int_{7}^{12} \frac{y}{y ^{2} - 3 y - 4} d y

\frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}

Decimal Notation

0.88176 \dots

Solution steps

\int_{7}^{12} \frac{y}{y ^{2} - 3 y - 4} d y

Take the partial fraction of

\frac{y}{y ^{2} - 3 y - 4} : \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )}

= \int_{7}^{12} \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )} 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_{7}^{12} \frac{1}{5 ( y + 1 )} d y + \int_{7}^{12} \frac{4}{5 ( y - 4 )} d y

\int_{7}^{12} \frac{1}{5 ( y + 1 )} d y = \frac{1}{5} (ln (13) - 3 ln (2))

\int_{7}^{12} \frac{4}{5 ( y - 4 )} d y = \frac{4}{5} (3 ln (2) - ln (3))

= \frac{1}{5} (ln (13) - 3 ln (2)) + \frac{4}{5} (3 ln (2) - ln (3))

Simplify

\frac{1}{5} (ln (13) - 3 ln (2)) + \frac{4}{5} (3 ln (2) - ln (3)) : \frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}

= \frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}

\int x sin (2 x) d x

\frac{d}{d x} (\frac{x}{y} + e^{y})

\int 5^{x} - sec^{2} (x) d x

- x^{2} - 1, x^{2} + 4, [- 2, 3]

What is the integral from 7 to 12 of y/(y^2-3y-4) ?
The integral from 7 to 12 of y/(y^2-3y-4) is (ln(13)+9ln(2)-4ln(3))/5