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

The integral from 2 to 4 of (x+2)/(x^2+3x-4) is ln(2)-1/10 ln(4/9)

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

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

ln (2) - \frac{1}{10} ln (\frac{4}{9})

Decimal Notation

0.77424 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{\frac{7}{2}}^{\frac{11}{2}} \frac{2 ( 2 u + 1 )}{4 u ^{2} - 25} d u

= \int_{\frac{7}{2}}^{\frac{11}{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

= 2 \cdot \int_{\frac{7}{2}}^{\frac{11}{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

= 2 (\int_{\frac{7}{2}}^{\frac{11}{2}} \frac{2 u}{4 u ^{2} - 25} d u + \int_{\frac{7}{2}}^{\frac{11}{2}} \frac{1}{4 u ^{2} - 25} d u)

\int_{\frac{7}{2}}^{\frac{11}{2}} \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{2} ln (2)

\int_{\frac{7}{2}}^{\frac{11}{2}} \frac{1}{4 u ^{2} - 25} d u = - \frac{ln ( \frac{4}{9} )}{20}

= 2 (\frac{1}{2} ln (2) - \frac{ln ( \frac{4}{9} )}{20})

Simplify

2 (\frac{1}{2} ln (2) - \frac{ln ( \frac{4}{9} )}{20}) : ln (2) - \frac{1}{10} ln (\frac{4}{9})

= ln (2) - \frac{1}{10} ln (\frac{4}{9})

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What is the integral from 2 to 4 of (x+2)/(x^2+3x-4) ?
The integral from 2 to 4 of (x+2)/(x^2+3x-4) is ln(2)-1/10 ln(4/9)