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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int cos (u) d u

= 3 \cdot \int cos (u) d u

Use the common integral:

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

= 3 sin (u)

Substitute back

u = arctan (x)

= 3 sin (arctan (x))

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 u}{1 + u ^{2}} d u

y^{^{''}} + y^{^{'}} - 2 y = 2 x

\int 2 e^{u} d u

\frac{( 3 y ^{2} - x ^{2} )}{y ^{5}} y^{^{'}} + \frac{x}{2 y ^{4}} = 0

f (x) = - 5.7 + 3.57 ln (x)

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