What is the integral from-1 to 4 of (7y)/(y^2-3y-10) ?

The integral from-1 to 4 of (7y)/(y^2-3y-10) is -3ln(6)

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integral from-1 to 4 of (7y)/(y^2-3y-10)

\int_{- 1}^{4} \frac{7 y}{y ^{2} - 3 y - 10} d y

- 3 ln (6)

Decimal Notation

- 5.37527 \dots

Solution steps

\int_{- 1}^{4} \frac{7 y}{y ^{2} - 3 y - 10} d y

Take the constant out:

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

= 7 \cdot \int_{- 1}^{4} \frac{y}{y ^{2} - 3 y - 10} d y

Take the partial fraction of

\frac{y}{y ^{2} - 3 y - 10} : \frac{2}{7 ( y + 2 )} + \frac{5}{7 ( y - 5 )}

= 7 \cdot \int_{- 1}^{4} \frac{2}{7 ( y + 2 )} + \frac{5}{7 ( y - 5 )} 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

= 7 (\int_{- 1}^{4} \frac{2}{7 ( y + 2 )} d y + \int_{- 1}^{4} \frac{5}{7 ( y - 5 )} d y)

\int_{- 1}^{4} \frac{2}{7 ( y + 2 )} d y = \frac{2}{7} ln (6)

\int_{- 1}^{4} \frac{5}{7 ( y - 5 )} d y = - \frac{5}{7} ln (6)

= 7 (\frac{2}{7} ln (6) - \frac{5}{7} ln (6))

Simplify

7 (\frac{2}{7} ln (6) - \frac{5}{7} ln (6)) : - 3 ln (6)

= - 3 ln (6)

x \to \frac{π}{2} + lim (sin (x))

\frac{d}{d x} ((3 x^{2} - 9)^{- 14})

y = x^{2} - 7 x + 7, (0, 7)

\int \frac{1}{x} cos^{2} (x) d x

\int_{0}^{a} 7 x^{2} a^{2} - x^{2} d x

What is the integral from-1 to 4 of (7y)/(y^2-3y-10) ?
The integral from-1 to 4 of (7y)/(y^2-3y-10) is -3ln(6)