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

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

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

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

\frac{1}{2} ln x^{2} + 4 x + 49 + C

Solution steps

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

= \frac{1}{2} ln x^{2} + 4 x + 49

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 49 + C

\frac{12}{6 x ^{2} + 1}

\int \frac{1}{( x ^{2} + 4 ) ^{\frac{3}{2}}} d x

\int_{0}^{4} 6 x - x^{2} d x

\frac{d y}{d x} = (x - 5) e^{- 2 y}, y (5) = ln (5)

\int \frac{3 x ^{2}}{49 - x ^{2}} d x

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