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

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

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

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

- 2 ln (2)

Decimal

- 1.38629 \dots

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

= 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)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4

\int csc^{3} (x) cot^{2} (x) d x

e^{θ}

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

y^{^{'}} = \frac{1}{y}, y (0) = 1

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