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

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

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

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

- 2 ln (2)

Decimal

- 1.38629 \dots

Solution steps

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

Take the constant out:

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

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

Complete the square

y^{2} - y - 6 : (y - \frac{1}{2})^{2} - \frac{25}{4}

= 5 \cdot \int_{- 1}^{2} \frac{y}{( y - \frac{1}{2} ) ^{2} - \frac{25}{4}} d y

Apply u-substitution

= 5 \cdot \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{2 u + 1}{4 u ^{2} - 25} d u

Expand

\frac{2 u + 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{1}{4 u ^{2} - 25}

Apply the Sum Rule:

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

= 5 \cdot 2 (\int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 25} d u + \int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 25} d u)

\int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 25} d u = 0

\int_{- \frac{3}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{5} ln (2)

= 5 \cdot 2 (0 - \frac{1}{5} ln (2))

Simplify

5 \cdot 2 (0 - \frac{1}{5} ln (2)) : - 2 ln (2)

= - 2 ln (2)

f (x) = x^{3} - 6 x^{2} + 9 x + 5

f (x) = x^{2} - 3 x + 6, at x = 2

7 x^{2}, x^{2} + 4

x \to \infty lim (x - x)

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

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