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

The integral of 3/(x^2+10x+29) is 3/2 arctan(1/2 (x+5))+C

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} + 10 x + 29} d x

Complete the square

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

= 3 \cdot \int \frac{1}{( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= 3 \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{2} arctan (v)

Substitute back

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

Simplify

3 \cdot \frac{1}{2} arctan (\frac{x + 5}{2}) : \frac{3}{2} arctan (\frac{1}{2} (x + 5))

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

Add a constant to the solution

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

x \to - 1 lim (\frac{1}{x ^{2} - 2 x - 3})

\frac{\partial}{\partial y} (\frac{x}{y ^{2} + 1})

\frac{\partial}{\partial x} (x^{3} + 4 x^{2} y + 4 x + 2)

x \to 2 - lim (\frac{2}{2 - x})

x \to 0 lim (2 - x^{2})

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