What is the integral of 4/(x^2+5x-4) ?

The integral of 4/(x^2+5x-4) is -4/(sqrt(41))(ln|(2x+5)/(sqrt(41))+1|-ln|(2x+5)/(sqrt(41))-1|)+C

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integral of 4/(x^2+5x-4)

\int \frac{4}{x ^{2} + 5 x - 4} d x

- \frac{4}{41} (ln \frac{2 x + 5}{41} + 1 - ln \frac{2 x + 5}{41} - 1) + C

Solution steps

\int \frac{4}{x ^{2} + 5 x - 4} d x

Take the constant out:

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

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

Complete the square

x^{2} + 5 x - 4 : (x + \frac{5}{2})^{2} - \frac{41}{4}

= 4 \cdot \int \frac{1}{( x + \frac{5}{2} ) ^{2} - \frac{41}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 41} d u

= 4 \cdot \int \frac{4}{4 u ^{2} - 41} d u

Take the constant out:

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

= 4 \cdot 4 \cdot \int \frac{1}{4 u ^{2} - 41} d u

Apply Integral Substitution

= 4 \cdot 4 \cdot \int \frac{1}{2 41 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 4 \cdot 4 \cdot \frac{1}{2 41} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 4 \cdot 4 \cdot \frac{1}{2 41} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot 4 \cdot \frac{1}{2 41} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 \cdot 4 \cdot \frac{1}{2 41} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 4 \cdot 4 \cdot \frac{1}{2 41} - \frac{ln \frac{2}{41} ( x + \frac{5}{2} ) + 1}{2} - \frac{ln \frac{2}{41} ( x + \frac{5}{2} ) - 1}{2}

Simplify

4 \cdot 4 \cdot \frac{1}{2 41} - \frac{ln \frac{2}{41} ( x + \frac{5}{2} ) + 1}{2} - \frac{ln \frac{2}{41} ( x + \frac{5}{2} ) - 1}{2} : - \frac{4}{41} (ln \frac{2 x + 5}{41} + 1 - ln \frac{2 x + 5}{41} - 1)

= - \frac{4}{41} (ln \frac{2 x + 5}{41} + 1 - ln \frac{2 x + 5}{41} - 1)

Add a constant to the solution

= - \frac{4}{41} (ln \frac{2 x + 5}{41} + 1 - ln \frac{2 x + 5}{41} - 1) + C

n = 1 \sum \infty \frac{9}{n ( n + 3 )}

\frac{- 1}{x}

\int x 8^{x} d x

\frac{\partial}{\partial x} (x^{\frac{1}{2}} y^{\frac{1}{2}})

\frac{3}{s ^{4}}

What is the integral of 4/(x^2+5x-4) ?
The integral of 4/(x^2+5x-4) is -4/(sqrt(41))(ln|(2x+5)/(sqrt(41))+1|-ln|(2x+5)/(sqrt(41))-1|)+C