What is the integral of (x+4)/(x^2+8x-20) ?

The integral of (x+4)/(x^2+8x-20) is 1/2 ln|x^2+8x-20|+C

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integral of (x+4)/(x^2+8x-20)

\int \frac{x + 4}{x ^{2} + 8 x - 20} d x

\frac{1}{2} ln x^{2} + 8 x - 20 + C

Solution steps

\int \frac{x + 4}{x ^{2} + 8 x - 20} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 8 x - 20

= \frac{1}{2} ln x^{2} + 8 x - 20

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 8 x - 20 + C

y^{^{'}} = 2 x y - x

x \to 0 lim (10 - x^{2})

\int \frac{sec ^{2} ( x )}{3} d x

\int_{0}^{1} 2^{x} d x

t + 1

What is the integral of (x+4)/(x^2+8x-20) ?
The integral of (x+4)/(x^2+8x-20) is 1/2 ln|x^2+8x-20|+C