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

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

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

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

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

Solution steps

\int \frac{x}{( x ^{2} + a ^{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 = x^{2} + a^{2}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 6 y^{^{'}} + 12 y = sin (2 x)

y^{^{''}} + y = e^{- x}

y = x (49.7)

\int \frac{1}{e ^{- 2 x}} d x

\int 2 x^{2} sin (π x) d x

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