What is the integral of x/((1+x^2)^{3/2)} ?

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

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integral of x/((1+x^2)^{3/2)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{3}{2}}} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

= \frac{1}{2} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = 1 + x^{2}

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

Simplify

\frac{1}{2} (- \frac{2}{( 1 + x ^{2} ) ^{\frac{1}{2}}}) : - \frac{1}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

x \to - 2 lim (62 \cdot x^{5})

f (x) = 3 sin (x - 2)

\int \frac{9}{4 - x ^{2}} d x

x \to 0 lim (2 - x)

x^{\frac{4}{3}}

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