What is the integral of 1/((x^2-4x+5)^2) ?

The integral of 1/((x^2-4x+5)^2) is 1/2 (arctan(x-2)+1/2 sin(2arctan(x-2)))+C

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integral of 1/((x^2-4x+5)^2)

\int \frac{1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

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

Solution steps

\int \frac{1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

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

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 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Add a constant to the solution

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

f (x) = 3 x^{2} + x - 2, at x = 1

\int \frac{x - 2}{x ^{2} - x + 1} d x

\frac{\partial}{\partial x} (7 e^{x y} ln (x^{2} + y^{2}))

\frac{d}{d x} (x^{- \frac{5}{3}})

\frac{1}{2 s + 4}

What is the integral of 1/((x^2-4x+5)^2) ?
The integral of 1/((x^2-4x+5)^2) is 1/2 (arctan(x-2)+1/2 sin(2arctan(x-2)))+C