What is the integral from-1 to 6 of (9y)/(y^2-5y-14) ?

The integral from-1 to 6 of (9y)/(y^2-5y-14) is -15ln(2)

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integral from-1 to 6 of (9y)/(y^2-5y-14)

\int_{- 1}^{6} \frac{9 y}{y ^{2} - 5 y - 14} d y

- 15 ln (2)

Decimal

- 10.39720 \dots

Solution steps

\int_{- 1}^{6} \frac{9 y}{y ^{2} - 5 y - 14} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{- 1}^{6} \frac{y}{y ^{2} - 5 y - 14} d y

Take the partial fraction of

\frac{y}{y ^{2} - 5 y - 14} : \frac{2}{9 ( y + 2 )} + \frac{7}{9 ( y - 7 )}

= 9 \cdot \int_{- 1}^{6} \frac{2}{9 ( y + 2 )} + \frac{7}{9 ( y - 7 )} 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

= 9 (\int_{- 1}^{6} \frac{2}{9 ( y + 2 )} d y + \int_{- 1}^{6} \frac{7}{9 ( y - 7 )} d y)

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

\int_{- 1}^{6} \frac{7}{9 ( y - 7 )} d y = - \frac{7}{3} ln (2)

= 9 (\frac{2}{3} ln (2) - \frac{7}{3} ln (2))

Simplify

9 (\frac{2}{3} ln (2) - \frac{7}{3} ln (2)) : - 15 ln (2)

= - 15 ln (2)

\int_{1}^{2} (\frac{4}{x} - \frac{16}{x ^{2}}) d x

y^{^{'}} - 2 t y = t

\int \frac{5 cos ^{2} ( x )}{sin ( x )} d x

t an g e n t y = \frac{8 x}{( x ^{2} + 1 )}, at

\int sec (6 x) tan (6 x) d x

What is the integral from-1 to 6 of (9y)/(y^2-5y-14) ?
The integral from-1 to 6 of (9y)/(y^2-5y-14) is -15ln(2)