What is the integral of (x+16)/(x^2+22x+125) ?

The integral of (x+16)/(x^2+22x+125) is 1/2 ln|x^2+22x+125|+5/2 arctan((x+11)/2)+C

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integral of (x+16)/(x^2+22x+125)

\int \frac{x + 16}{x ^{2} + 22 x + 125} d x

\frac{1}{2} ln x^{2} + 22 x + 125 + \frac{5}{2} arctan (\frac{x + 11}{2}) + C

Solution steps

\int \frac{x + 16}{x ^{2} + 22 x + 125} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 10 u + 29} d u

= \int \frac{u}{u ^{2} - 10 u + 29} d u

Complete the square

u^{2} - 10 u + 29 : (u - 5)^{2} + 4

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

Apply u-substitution

: \int \frac{v + 5}{v ^{2} + 4} d v

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

Expand

\frac{v + 5}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{5}{v ^{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{v}{v ^{2} + 4} d v + \int \frac{5}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

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

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

Substitute back

= \frac{1}{2} ln (x + 16 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 16 - 5}{2})

Simplify

\frac{1}{2} ln (x + 16 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 16 - 5}{2}) : \frac{1}{2} ln x^{2} + 22 x + 125 + \frac{5}{2} arctan (\frac{x + 11}{2})

= \frac{1}{2} ln x^{2} + 22 x + 125 + \frac{5}{2} arctan (\frac{x + 11}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 22 x + 125 + \frac{5}{2} arctan (\frac{x + 11}{2}) + C

\int 9 cos^{2} (x t a) n^{3} x d x

\frac{\partial}{\partial x} (ln (x + 2 y + 3 z))

x \to + 0 + lim ((1 + x)^{\frac{2}{x}})

\frac{( 2 + x )}{1 - x}

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

What is the integral of (x+16)/(x^2+22x+125) ?
The integral of (x+16)/(x^2+22x+125) is 1/2 ln|x^2+22x+125|+5/2 arctan((x+11)/2)+C