What is the integral of (x+3)/(x^2+6x+7) ?

The integral of (x+3)/(x^2+6x+7) is 1/2 ln|x^2+6x+7|+C

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integral of (x+3)/(x^2+6x+7)

\int \frac{x + 3}{x ^{2} + 6 x + 7} d x

\frac{1}{2} ln x^{2} + 6 x + 7 + C

Solution steps

\int \frac{x + 3}{x ^{2} + 6 x + 7} d x

Apply u-substitution

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

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

Apply u-substitution

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

= \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 + 3)^{2} - 2

Simplify

\frac{1}{2} ln (x + 3)^{2} - 2 : \frac{1}{2} ln x^{2} + 6 x + 7

= \frac{1}{2} ln x^{2} + 6 x + 7

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 7 + C

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

\int e^{2 x} (cos (3 x)) d x

\int \frac{6}{cos ( x ) - 1} d x

2 y^{^{''}} + 3 y^{^{'}} - 2 y = cos (2 x)

\frac{x}{x ^{2}}

What is the integral of (x+3)/(x^2+6x+7) ?
The integral of (x+3)/(x^2+6x+7) is 1/2 ln|x^2+6x+7|+C