What is the integral of 1/(sqrt((x^2+4x+7)^3)) ?

The integral of 1/(sqrt((x^2+4x+7)^3)) is (x+2)/(3sqrt(x^2+4x+7))+C

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integral of 1/(sqrt((x^2+4x+7)^3))

\int \frac{1}{( x ^{2} + 4 x + 7 ) ^{3}} d x

\frac{x + 2}{3 x ^{2} + 4 x + 7} + C

Solution steps

\int \frac{1}{( x ^{2} + 4 x + 7 ) ^{3}} d x

(x^{2} + 4 x + 7)^{3} = (x^{2} + 4 x + 7)^{\frac{3}{2}},

assuming

(x^{2} + 4 x + 7) \geq 0

= \int \frac{1}{( x ^{2} + 4 x + 7 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 4 x + 7 : (x + 2)^{2} + 3

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

Apply u-substitution

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

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{3} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{3} sin (v)

Substitute back

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

Simplify

\frac{1}{3} sin (arctan (\frac{1}{3} (x + 2))) : \frac{x + 2}{3 x ^{2} + 4 x + 7}

= \frac{x + 2}{3 x ^{2} + 4 x + 7}

Add a constant to the solution

= \frac{x + 2}{3 x ^{2} + 4 x + 7} + C

What is the integral of 1/(sqrt((x^2+4x+7)^3)) ?
The integral of 1/(sqrt((x^2+4x+7)^3)) is (x+2)/(3sqrt(x^2+4x+7))+C