What is the integral of (6x+24)/(x^2+8x-10) ?

The integral of (6x+24)/(x^2+8x-10) is 3ln|x^2+8x-10|+C

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integral of (6x+24)/(x^2+8x-10)

\int \frac{6 x + 24}{x ^{2} + 8 x - 10} d x

3 ln x^{2} + 8 x - 10 + C

Solution steps

\int \frac{6 x + 24}{x ^{2} + 8 x - 10} d x

Complete the square

x^{2} + 8 x - 10 : (x + 4)^{2} - 26

= \int \frac{6 x + 24}{( x + 4 ) ^{2} - 26} d x

Apply u-substitution

: \int \frac{6 u}{u ^{2} - 26} d u

= \int \frac{6 u}{u ^{2} - 26} d u

Take the constant out:

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

= 6 \cdot \int \frac{u}{u ^{2} - 26} d u

Apply u-substitution

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

= 6 \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 6 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= 6 \cdot \frac{1}{2} ln (x + 4)^{2} - 26

Simplify

6 \cdot \frac{1}{2} ln (x + 4)^{2} - 26 : 3 ln x^{2} + 8 x - 10

= 3 ln x^{2} + 8 x - 10

Add a constant to the solution

= 3 ln x^{2} + 8 x - 10 + C

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What is the integral of (6x+24)/(x^2+8x-10) ?
The integral of (6x+24)/(x^2+8x-10) is 3ln|x^2+8x-10|+C