What is the integral of (x+2)/(x^2+7x-8) ?

The integral of (x+2)/(x^2+7x-8) is 2(1/4 ln|4x^2+28x-32|+1/12 (ln| 16/9+2/9 x|-ln|-2/9+2/9 x|))+C

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integral of (x+2)/(x^2+7x-8)

\int \frac{x + 2}{x ^{2} + 7 x - 8} d x

2 (\frac{1}{4} ln 4 x^{2} + 28 x - 32 + \frac{1}{12} (ln \frac{16}{9} + \frac{2}{9} x - ln - \frac{2}{9} + \frac{2}{9} x)) + C

Solution steps

\int \frac{x + 2}{x ^{2} + 7 x - 8} d x

Complete the square

x^{2} + 7 x - 8 : (x + \frac{7}{2})^{2} - \frac{81}{4}

= \int \frac{x + 2}{( x + \frac{7}{2} ) ^{2} - \frac{81}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 3 )}{4 u ^{2} - 81} d u

= \int \frac{2 ( 2 u - 3 )}{4 u ^{2} - 81} 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 - 3}{4 u ^{2} - 81} d u

Expand

\frac{2 u - 3}{4 u ^{2} - 81} : \frac{2 u}{4 u ^{2} - 81} - \frac{3}{4 u ^{2} - 81}

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} - 81} d u - \int \frac{3}{4 u ^{2} - 81} d u)

\int \frac{2 u}{4 u ^{2} - 81} d u = \frac{1}{4} ln 4 u^{2} - 81

\int \frac{3}{4 u ^{2} - 81} d u = - \frac{1}{12} (ln \frac{2}{9} u + 1 - ln \frac{2}{9} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 81 - (- \frac{1}{12} (ln \frac{2}{9} u + 1 - ln \frac{2}{9} u - 1)))

Substitute back

u = x + \frac{7}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 81 - (- \frac{1}{12} (ln \frac{2}{9} (x + \frac{7}{2}) + 1 - ln \frac{2}{9} (x + \frac{7}{2}) - 1)))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 81 - (- \frac{1}{12} (ln \frac{2}{9} (x + \frac{7}{2}) + 1 - ln \frac{2}{9} (x + \frac{7}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 28 x - 32 + \frac{1}{12} (ln \frac{16}{9} + \frac{2}{9} x - ln - \frac{2}{9} + \frac{2}{9} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x - 32 + \frac{1}{12} (ln \frac{16}{9} + \frac{2}{9} x - ln - \frac{2}{9} + \frac{2}{9} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x - 32 + \frac{1}{12} (ln \frac{16}{9} + \frac{2}{9} x - ln - \frac{2}{9} + \frac{2}{9} x)) + C

2.87 y^{^{''}} + 10.5 y^{^{'}} - 1.83 y = 0

y^{^{'}} = y^{2} (1 + e^{x})

\frac{d}{d x} (\frac{4 x ^{5} + 7}{x ^{3}})

y^{^{''}} - 2 y^{^{'}} + y = \frac{( - 6 e ^{t} )}{t ^{2} + 1}

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = 2

What is the integral of (x+2)/(x^2+7x-8) ?
The integral of (x+2)/(x^2+7x-8) is 2(1/4 ln|4x^2+28x-32|+1/12 (ln| 16/9+2/9 x|-ln|-2/9+2/9 x|))+C