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

\int \frac{1}{( x + 4 ) ( x + 8 )} d x

(3, - 2), (7, 0)

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

(3.12) (5.6)

y^{^{''}} + y = 7 e^{- 5 x}, y (0) = 0, y^{^{'}} (0) = 0

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