What is the integral of (2x-3)/(x^2+10x+29) ?

The integral of (2x-3)/(x^2+10x+29) is ln| 1/4 (x+5)^2+1|-13/2 arctan(1/2 (x+5))+C

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integral of (2x-3)/(x^2+10x+29)

\int \frac{2 x - 3}{x ^{2} + 10 x + 29} d x

ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5)) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 10 x + 29} d x

Complete the square

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

= \int \frac{2 x - 3}{( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{2 u - 13}{u ^{2} + 4} d u

= \int \frac{2 u - 13}{u ^{2} + 4} d u

Apply Integral Substitution

= \int \frac{4 v - 13}{2 ( v ^{2} + 1 )} 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{4 v - 13}{v ^{2} + 1} d v

Expand

\frac{4 v - 13}{v ^{2} + 1} : \frac{4 v}{v ^{2} + 1} - \frac{13}{v ^{2} + 1}

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{4 v}{v ^{2} + 1} d v - \int \frac{13}{v ^{2} + 1} d v)

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

\int \frac{13}{v ^{2} + 1} d v = 13 arctan (v)

= \frac{1}{2} (2 ln v^{2} + 1 - 13 arctan (v))

Substitute back

= \frac{1}{2} (2 ln (\frac{x + 5}{2})^{2} + 1 - 13 arctan (\frac{x + 5}{2}))

Simplify

\frac{1}{2} (2 ln (\frac{x + 5}{2})^{2} + 1 - 13 arctan (\frac{x + 5}{2})) : ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5))

= ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5))

Add a constant to the solution

= ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5)) + C

\frac{d}{d x} (x + cot (\frac{x}{2}))

\frac{\partial}{\partial y} (10 x y e^{- x^{2} - y^{2}})

\frac{d}{d x} (- 3 cot (1 - x^{2}))

x \to 0 lim (3 - \frac{1}{x})

f (x) = π^{x}

What is the integral of (2x-3)/(x^2+10x+29) ?
The integral of (2x-3)/(x^2+10x+29) is ln| 1/4 (x+5)^2+1|-13/2 arctan(1/2 (x+5))+C