What is the integral from 6 to 9 of y/(y^2-2y-3) ?

The integral from 6 to 9 of y/(y^2-2y-3) is (ln(10)-ln(7)+3ln(2))/4

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integral from 6 to 9 of y/(y^2-2y-3)

\int_{6}^{9} \frac{y}{y ^{2} - 2 y - 3} d y

\frac{ln ( 10 ) - ln ( 7 ) + 3 ln ( 2 )}{4}

Decimal Notation

0.60902 \dots

Solution steps

\int_{6}^{9} \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}^{9} \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}^{9} \frac{1}{4 ( y + 1 )} d y + \int_{6}^{9} \frac{3}{4 ( y - 3 )} d y

\int_{6}^{9} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (ln (10) - ln (7))

\int_{6}^{9} \frac{3}{4 ( y - 3 )} d y = \frac{3}{4} ln (2)

= \frac{1}{4} (ln (10) - ln (7)) + \frac{3}{4} ln (2)

Simplify

\frac{1}{4} (ln (10) - ln (7)) + \frac{3}{4} ln (2) : \frac{ln ( 10 ) - ln ( 7 ) + 3 ln ( 2 )}{4}

= \frac{ln ( 10 ) - ln ( 7 ) + 3 ln ( 2 )}{4}

\frac{d}{d x} (3 csc (x) + 5 cos (x))

\int_{4}^{\infty} x e^{- 4 x} d x

\frac{\partial}{\partial x} (3 x y ln (x y))

x^{4} - y^{2} = 7, (2, 3)

\int sec^{2} (θ) tan^{9} (θ) d θ

What is the integral from 6 to 9 of y/(y^2-2y-3) ?
The integral from 6 to 9 of y/(y^2-2y-3) is (ln(10)-ln(7)+3ln(2))/4