What is the integral of (5x)/((x+3)(x-2)) ?

The integral of (5x)/((x+3)(x-2)) is 10(1/4 ln|4x^2+4x-24|+1/20 (ln| 6/5+2/5 x|-ln|-4/5+2/5 x|))+C

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integral of (5x)/((x+3)(x-2))

\int \frac{5 x}{( x + 3 ) ( x - 2 )} d x

10 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

Solution steps

\int \frac{5 x}{( x + 3 ) ( x - 2 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{( x + 3 ) ( x - 2 )} d x

Complete the square

(x + 3) (x - 2) : (x + \frac{1}{2})^{2} - \frac{25}{4}

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

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 25} d u

= 5 \cdot \int \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{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{2 u}{4 u ^{2} - 25} d u - \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 5 \cdot 2 (\frac{1}{4} ln 4 u^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

Substitute back

u = x + \frac{1}{2}

= 5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

Simplify

5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 10 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

= 10 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

Add a constant to the solution

= 10 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

\int u^{- 5} (5 - u^{5} + u^{10}) d u

\int x^{18} d x

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 5

\int_{0}^{2 π} 5 d x

x \to 0 + lim (- x ln (x))

What is the integral of (5x)/((x+3)(x-2)) ?
The integral of (5x)/((x+3)(x-2)) is 10(1/4 ln|4x^2+4x-24|+1/20 (ln| 6/5+2/5 x|-ln|-4/5+2/5 x|))+C