What is the integral of x/((sqrt(16+x^2))^3) ?

The integral of x/((sqrt(16+x^2))^3) is (x^2)/(16sqrt(16+x^2))-1/16 sqrt(16+x^2)+C

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integral of x/((sqrt(16+x^2))^3)

\int \frac{x}{( 16 + x ^{2} ) ^{3}} d x

\frac{x ^{2}}{16 16 + x ^{2}} - \frac{1}{16} 16 + x^{2} + C

Solution steps

\int \frac{x}{( 16 + x ^{2} ) ^{3}} d x

Apply exponent rule:

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

(16 + x^{2})^{3} = (16 + x^{2})^{\frac{3}{2}}

= \int \frac{x}{( 16 + x ^{2} ) ^{\frac{3}{2}}} d x

16 + x^{2} = 16 (1 + \frac{x ^{2}}{16})

= \int \frac{x}{( 16 ( 1 + \frac{x ^{2}}{16} ) ) ^{\frac{3}{2}}} d x

Apply Integration By Parts

: \frac{x ^{2}}{16 16 + x ^{2}} - \int \frac{x}{16 16 + x ^{2}} d x

= \frac{x ^{2}}{16 16 + x ^{2}} - \int \frac{x}{16 16 + x ^{2}} d x

\int \frac{x}{16 16 + x ^{2}} d x = \frac{1}{16} 16 + x^{2}

= \frac{x ^{2}}{16 16 + x ^{2}} - \frac{1}{16} 16 + x^{2}

Add a constant to the solution

= \frac{x ^{2}}{16 16 + x ^{2}} - \frac{1}{16} 16 + x^{2} + C

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What is the integral of x/((sqrt(16+x^2))^3) ?
The integral of x/((sqrt(16+x^2))^3) is (x^2)/(16sqrt(16+x^2))-1/16 sqrt(16+x^2)+C