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

The integral of 1/((x^2+2x+4)^{5/2)} is (3(x+1)(x^2+2x+4)-(x+1)^3)/(27(x^2+2x+4)^{3/2)}+C

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

integral

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

\frac{3 ( x + 1 ) ( x ^{2} + 2 x + 4 ) - ( x + 1 ) ^{3}}{27 ( x ^{2} + 2 x + 4 ) ^{\frac{3}{2}}} + C

Solution steps

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

Complete the square

x^{2} + 2 x + 4 : (x + 1)^{2} + 3

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

Apply u-substitution

= \int \frac{1}{( u ^{2} + 3 ) ^{\frac{5}{2}}} d u

Apply Trigonometric Substitution

= \int \frac{1}{9 sec ^{3} ( v )} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sec ^{3} ( v )} d v

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 - sin^{2} (v)) cos (v) d v

Apply u-substitution

= \frac{1}{9} \cdot \int 1 - w^{2} 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}{9} (\int 1 d w - \int w^{2} d w)

\int 1 d w = w

\int w^{2} d w = \frac{w ^{3}}{3}

= \frac{1}{9} (w - \frac{w ^{3}}{3})

Substitute back

= \frac{1}{9} sin (arctan (\frac{1}{3 ^{\frac{1}{2}}} (x + 1))) - \frac{sin ^{3} ( arctan ( \frac{1}{3 ^{\frac{1}{2}}} ( x + 1 ) ) )}{3}

Simplify

\frac{1}{9} sin (arctan (\frac{1}{3 ^{\frac{1}{2}}} (x + 1))) - \frac{sin ^{3} ( arctan ( \frac{1}{3 ^{\frac{1}{2}}} ( x + 1 ) ) )}{3} : \frac{3 ( x + 1 ) ( x ^{2} + 2 x + 4 ) - ( x + 1 ) ^{3}}{27 ( x ^{2} + 2 x + 4 ) ^{\frac{3}{2}}}

= \frac{3 ( x + 1 ) ( x ^{2} + 2 x + 4 ) - ( x + 1 ) ^{3}}{27 ( x ^{2} + 2 x + 4 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{3 ( x + 1 ) ( x ^{2} + 2 x + 4 ) - ( x + 1 ) ^{3}}{27 ( x ^{2} + 2 x + 4 ) ^{\frac{3}{2}}} + C

What is the integral of 1/((x^2+2x+4)^{5/2)} ?
The integral of 1/((x^2+2x+4)^{5/2)} is (3(x+1)(x^2+2x+4)-(x+1)^3)/(27(x^2+2x+4)^{3/2)}+C