What is the integral from 0 to 4 of (8y-1)/(2y^2-15y-8) ?

The integral from 0 to 4 of (8y-1)/(2y^2-15y-8) is (4ln(9/65536)+ln(18))/(17)

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integral from 0 to 4 of (8y-1)/(2y^2-15y-8)

\int_{0}^{4} \frac{8 y - 1}{2 y ^{2} - 15 y - 8} d y

\frac{4 ln ( \frac{9}{65536} ) + ln ( 18 )}{17}

Decimal Notation

- 1.92247 \dots

Solution steps

\int_{0}^{4} \frac{8 y - 1}{2 y ^{2} - 15 y - 8} d y

Expand

\frac{8 y - 1}{2 y ^{2} - 15 y - 8} : \frac{8 y}{2 y ^{2} - 15 y - 8} - \frac{1}{2 y ^{2} - 15 y - 8}

Apply the Sum Rule:

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

= \int_{0}^{4} \frac{8 y}{2 y ^{2} - 15 y - 8} d y - \int_{0}^{4} \frac{1}{2 y ^{2} - 15 y - 8} d y

\int_{0}^{4} \frac{8 y}{2 y ^{2} - 15 y - 8} d y = \frac{4 ln ( \frac{9}{65536} )}{17}

\int_{0}^{4} \frac{1}{2 y ^{2} - 15 y - 8} d y = - \frac{ln ( 18 )}{17}

= \frac{4 ln ( \frac{9}{65536} )}{17} - (- \frac{ln ( 18 )}{17})

Simplify

\frac{4 ln ( \frac{9}{65536} )}{17} - (- \frac{ln ( 18 )}{17}) : \frac{4 ln ( \frac{9}{65536} ) + ln ( 18 )}{17}

= \frac{4 ln ( \frac{9}{65536} ) + ln ( 18 )}{17}

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What is the integral from 0 to 4 of (8y-1)/(2y^2-15y-8) ?
The integral from 0 to 4 of (8y-1)/(2y^2-15y-8) is (4ln(9/65536)+ln(18))/(17)