What is the integral of (x+4)/(x^2+8x+49) ?

The integral of (x+4)/(x^2+8x+49) is 1/2 ln|x^2+8x+49|+C

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integral of (x+4)/(x^2+8x+49)

\int \frac{x + 4}{x ^{2} + 8 x + 49} d x

\frac{1}{2} ln x^{2} + 8 x + 49 + C

Solution steps

\int \frac{x + 4}{x ^{2} + 8 x + 49} d x

Apply u-substitution

= \int \frac{u}{u ^{2} + 33} d u

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (x + 4)^{2} + 33

Simplify

\frac{1}{2} ln (x + 4)^{2} + 33 : \frac{1}{2} ln x^{2} + 8 x + 49

= \frac{1}{2} ln x^{2} + 8 x + 49

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 8 x + 49 + C

What is the integral of (x+4)/(x^2+8x+49) ?
The integral of (x+4)/(x^2+8x+49) is 1/2 ln|x^2+8x+49|+C