What is the integral of (7x+14)/(x^2+5x-6) ?

The integral of (7x+14)/(x^2+5x-6) is

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integral of (7x+14)/(x^2+5x-6)

\int \frac{7 x + 14}{x ^{2} + 5 x - 6} d x

4 (7 (\frac{1}{8} ln 4 x^{2} + 20 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) - \frac{1}{2} ln \frac{12}{7} + \frac{2}{7} x + \frac{1}{2} ln - \frac{2}{7} + \frac{2}{7} x) + C

Solution steps

\int \frac{7 x + 14}{x ^{2} + 5 x - 6} d x

Complete the square

x^{2} + 5 x - 6 : (x + \frac{5}{2})^{2} - \frac{49}{4}

= \int \frac{7 x + 14}{( x + \frac{5}{2} ) ^{2} - \frac{49}{4}} d x

Apply u-substitution

: \int \frac{4 ( 7 ( u - \frac{5}{2} ) + 14 )}{4 u ^{2} - 49} d u

= \int \frac{4 ( 7 ( u - \frac{5}{2} ) + 14 )}{4 u ^{2} - 49} d u

Take the constant out:

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

= 4 \cdot \int \frac{7 ( u - \frac{5}{2} ) + 14}{4 u ^{2} - 49} d u

Expand

\frac{7 ( u - \frac{5}{2} ) + 14}{4 u ^{2} - 49} : \frac{7 ( u - \frac{5}{2} )}{4 u ^{2} - 49} + \frac{14}{4 u ^{2} - 49}

Apply the Sum Rule:

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

= 4 (\int \frac{7 ( u - \frac{5}{2} )}{4 u ^{2} - 49} d u + \int \frac{14}{4 u ^{2} - 49} d u)

\int \frac{7 ( u - \frac{5}{2} )}{4 u ^{2} - 49} d u = 7 (\frac{1}{8} ln 4 u^{2} - 49 + \frac{5}{56} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1))

\int \frac{14}{4 u ^{2} - 49} d u = - \frac{1}{2} ln \frac{2}{7} u + 1 + \frac{1}{2} ln \frac{2}{7} u - 1

= 4 (7 (\frac{1}{8} ln 4 u^{2} - 49 + \frac{5}{56} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)) - \frac{1}{2} ln \frac{2}{7} u + 1 + \frac{1}{2} ln \frac{2}{7} u - 1)

Substitute back

u = x + \frac{5}{2}

= 4 (7 (\frac{1}{8} ln 4 (x + \frac{5}{2})^{2} - 49 + \frac{5}{56} (ln \frac{2}{7} (x + \frac{5}{2}) + 1 - ln \frac{2}{7} (x + \frac{5}{2}) - 1)) - \frac{1}{2} ln \frac{2}{7} (x + \frac{5}{2}) + 1 + \frac{1}{2} ln \frac{2}{7} (x + \frac{5}{2}) - 1)

Simplify

4 (7 (\frac{1}{8} ln 4 (x + \frac{5}{2})^{2} - 49 + \frac{5}{56} (ln \frac{2}{7} (x + \frac{5}{2}) + 1 - ln \frac{2}{7} (x + \frac{5}{2}) - 1)) - \frac{1}{2} ln \frac{2}{7} (x + \frac{5}{2}) + 1 + \frac{1}{2} ln \frac{2}{7} (x + \frac{5}{2}) - 1) : 4 (7 (\frac{1}{8} ln 4 x^{2} + 20 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) - \frac{1}{2} ln \frac{12}{7} + \frac{2}{7} x + \frac{1}{2} ln - \frac{2}{7} + \frac{2}{7} x)

= 4 (7 (\frac{1}{8} ln 4 x^{2} + 20 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) - \frac{1}{2} ln \frac{12}{7} + \frac{2}{7} x + \frac{1}{2} ln - \frac{2}{7} + \frac{2}{7} x)

Add a constant to the solution

= 4 (7 (\frac{1}{8} ln 4 x^{2} + 20 x - 24 + \frac{5}{56} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) - \frac{1}{2} ln \frac{12}{7} + \frac{2}{7} x + \frac{1}{2} ln - \frac{2}{7} + \frac{2}{7} x) + C

\int \frac{1}{3 x ^{2} - 2 x + 4} d x

y^{^{''}} + y = sec (t) tan (t)

2 - x^{2}, x, - 2, 1

n = 1 \sum \infty \frac{n !}{n ! + 1}

\int_{0}^{1} (\frac{x}{x ^{2} + 1}) d x

What is the integral of (7x+14)/(x^2+5x-6) ?
The integral of (7x+14)/(x^2+5x-6) is