What is the integral of (x-11)/(x^2+3x-4) ?

The integral of (x-11)/(x^2+3x-4) is 2(1/4 ln|4x^2+12x-16|+5/4 (ln| 8/5+2/5 x|-ln|-2/5+2/5 x|))+C

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integral of (x-11)/(x^2+3x-4)

\int \frac{x - 11}{x ^{2} + 3 x - 4} d x

2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

Solution steps

\int \frac{x - 11}{x ^{2} + 3 x - 4} d x

Complete the square

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

= \int \frac{x - 11}{( x + \frac{3}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

Substitute back

u = x + \frac{3}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - (- \frac{5}{4} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1)))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} - 25 - (- \frac{5}{4} (ln \frac{2}{5} (x + \frac{3}{2}) + 1 - ln \frac{2}{5} (x + \frac{3}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 12 x - 16 + \frac{5}{4} (ln \frac{8}{5} + \frac{2}{5} x - ln - \frac{2}{5} + \frac{2}{5} x)) + C

n = 0 \sum \infty \frac{4 ^{n}}{n !}

x \to \infty lim (1. 1^{x})

x \to 1 lim (x^{4} - 3 x^{2} + 6)

x \to - 1 lim (\frac{x ^{3} + x}{1 + x})

\frac{\partial}{\partial x} (x \cdot y \cdot cos (3 x) + e^{x y})

What is the integral of (x-11)/(x^2+3x-4) ?
The integral of (x-11)/(x^2+3x-4) is 2(1/4 ln|4x^2+12x-16|+5/4 (ln| 8/5+2/5 x|-ln|-2/5+2/5 x|))+C