What is the integral of ((4x+10))/(x^2+2x+5) ?

The integral of ((4x+10))/(x^2+2x+5) is 2ln|x^2+2x+5|+3arctan(1/2 (x+1))+C

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integral of ((4x+10))/(x^2+2x+5)

\int \frac{( 4 x + 10 )}{x ^{2} + 2 x + 5} d x

2 ln x^{2} + 2 x + 5 + 3 arctan (\frac{1}{2} (x + 1)) + C

Solution steps

\int \frac{4 x + 10}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

= \int \frac{4 x + 10}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{4 u + 6}{u ^{2} + 4} d u

= \int \frac{4 u + 6}{u ^{2} + 4} d u

Expand

\frac{4 u + 6}{u ^{2} + 4} : \frac{4 u}{u ^{2} + 4} + \frac{6}{u ^{2} + 4}

Apply the Sum Rule:

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

= \int \frac{4 u}{u ^{2} + 4} d u + \int \frac{6}{u ^{2} + 4} d u

\int \frac{4 u}{u ^{2} + 4} d u = 2 ln u^{2} + 4

\int \frac{6}{u ^{2} + 4} d u = 3 arctan (\frac{u}{2})

= 2 ln u^{2} + 4 + 3 arctan (\frac{u}{2})

Substitute back

u = x + 1

= 2 ln (x + 1)^{2} + 4 + 3 arctan (\frac{x + 1}{2})

Simplify

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

= 2 ln x^{2} + 2 x + 5 + 3 arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= 2 ln x^{2} + 2 x + 5 + 3 arctan (\frac{1}{2} (x + 1)) + C

n = 1 \sum \infty \frac{4}{3 ^{n} + 1}

y = 3 x^{2} + 2^{2} x^{2} + 1, (- 1, 0)

\int \frac{- 1}{x + 4} d x

f (x) = 6 + ln (x), at x = 1

\frac{\partial}{\partial x} (e^{3 x} (x^{2} + y^{2}))

What is the integral of ((4x+10))/(x^2+2x+5) ?
The integral of ((4x+10))/(x^2+2x+5) is 2ln|x^2+2x+5|+3arctan(1/2 (x+1))+C