What is the integral of 1/((x^2-6x+13)^2) ?

The integral of 1/((x^2-6x+13)^2) is 1/16 (arctan(1/2 (x-3))+1/2 sin(2arctan(1/2 (x-3))))+C

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integral of 1/((x^2-6x+13)^2)

\int \frac{1}{( x ^{2} - 6 x + 13 ) ^{2}} d x

\frac{1}{16} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3)))) + C

Solution steps

\int \frac{1}{( x ^{2} - 6 x + 13 ) ^{2}} d x

Complete the square

x^{2} - 6 x + 13 : (x - 3)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{8 ( v ^{2} + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

= \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Rewrite using trig identities

= \frac{1}{8} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

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}{8} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= \frac{1}{8} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Substitute back

= \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x - 3}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 3}{2})))

Simplify

\frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x - 3}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 3}{2}))) : \frac{1}{16} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3))))

= \frac{1}{16} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3))))

Add a constant to the solution

= \frac{1}{16} (arctan (\frac{1}{2} (x - 3)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 3)))) + C

\int_{0}^{1} \frac{50}{4 y - 1} d y

{t + e^{2 t}}

x \cdot (y^{2} - 1) = y \cdot (4 - x^{2}) \cdot y^{^{'}}

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

\int x^{2} cos (1 - x^{3}) d x

What is the integral of 1/((x^2-6x+13)^2) ?
The integral of 1/((x^2-6x+13)^2) is 1/16 (arctan(1/2 (x-3))+1/2 sin(2arctan(1/2 (x-3))))+C