What is the integral of 9/((sqrt(4+x^2))^3) ?

The integral of 9/((sqrt(4+x^2))^3) is (9x)/(4sqrt(4+x^2))+C

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integral of 9/((sqrt(4+x^2))^3)

\int \frac{9}{( 4 + x ^{2} ) ^{3}} d x

\frac{9 x}{4 4 + x ^{2}} + C

Solution steps

\int \frac{9}{( 4 + x ^{2} ) ^{3}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( 4 + x ^{2} ) ^{3}} d x

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

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

= 9 \cdot \int \frac{1}{( 4 + x ^{2} ) ^{\frac{3}{2}}} d x

Multiply by the conjugate of

4 + x^{2} : \frac{16 - x ^{4}}{4 - x ^{2}}

= 9 \cdot \int \frac{1}{( \frac{16 - x ^{4}}{4 - x ^{2}} ) ^{\frac{3}{2}}} d x

Apply Integral Substitution

= 9 \cdot \int \frac{1}{4 ( u ^{2} + 1 ) ^{\frac{3}{2}}} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int cos (v) d v

= 9 \cdot \frac{1}{4} \cdot \int cos (v) d v

Use the common integral:

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

= 9 \cdot \frac{1}{4} sin (v)

Substitute back

= 9 \cdot \frac{1}{4} sin (arctan (\frac{x}{2}))

Simplify

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

= \frac{9 x}{4 4 + x ^{2}}

Add a constant to the solution

= \frac{9 x}{4 4 + x ^{2}} + C

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What is the integral of 9/((sqrt(4+x^2))^3) ?
The integral of 9/((sqrt(4+x^2))^3) is (9x)/(4sqrt(4+x^2))+C