What is the integral of (5(4x+2))/(2(x^2+4x+5)) ?

The integral of (5(4x+2))/(2(x^2+4x+5)) is 5(ln|x^2+4x+5|-3arctan(x+2))+C

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integral of (5(4x+2))/(2(x^2+4x+5))

\int \frac{5 ( 4 x + 2 )}{2 ( x ^{2} + 4 x + 5 )} d x

5 (ln x^{2} + 4 x + 5 - 3 arctan (x + 2)) + C

Solution steps

\int \frac{5 ( 4 x + 2 )}{2 ( x ^{2} + 4 x + 5 )} d x

Take the constant out:

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

= \frac{5}{2} \cdot \int \frac{4 x + 2}{x ^{2} + 4 x + 5} d x

Expand

\frac{4 x + 2}{x ^{2} + 4 x + 5} : \frac{4 x}{x ^{2} + 4 x + 5} + \frac{2}{x ^{2} + 4 x + 5}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{5}{2} (\int \frac{4 x}{x ^{2} + 4 x + 5} d x + \int \frac{2}{x ^{2} + 4 x + 5} d x)

\int \frac{4 x}{x ^{2} + 4 x + 5} d x = 4 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

\int \frac{2}{x ^{2} + 4 x + 5} d x = 2 arctan (x + 2)

= \frac{5}{2} (4 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + 2 arctan (x + 2))

Simplify

\frac{5}{2} (4 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + 2 arctan (x + 2)) : 5 (ln x^{2} + 4 x + 5 - 3 arctan (x + 2))

= 5 (ln x^{2} + 4 x + 5 - 3 arctan (x + 2))

Add a constant to the solution

= 5 (ln x^{2} + 4 x + 5 - 3 arctan (x + 2)) + C

f (x) = \frac{x ^{2}}{( 1 + x )}

n = 1 \sum \infty (1 - \frac{1}{n})^{n}

y = \frac{x}{2 - tan ( x )}

\int \frac{9 x + 7}{( 18 x ^{2} + 28 x ) ^{4}} d x

\int ln (x^{2} + 9) d x

What is the integral of (5(4x+2))/(2(x^2+4x+5)) ?
The integral of (5(4x+2))/(2(x^2+4x+5)) is 5(ln|x^2+4x+5|-3arctan(x+2))+C