What is the integral of 6/((x^2+1)^{3/2)} ?

The integral of 6/((x^2+1)^{3/2)} is (6x)/((1+x^2)^{1/2)}+C

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integral of 6/((x^2+1)^{3/2)}

\int \frac{6}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{6 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{6}{( 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

= 6 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= 6 \cdot \int cos (u) d u

Use the common integral:

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

= 6 sin (u)

Substitute back

u = arctan (x)

= 6 sin (arctan (x))

Simplify

6 sin (arctan (x)) : \frac{6 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{6 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{6 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

f (x) = 9 π x^{2}

\frac{d}{d x} (\frac{2 cos ( x ) + 3}{5 sin ( x ) - 2})

\frac{d}{d x} (- \frac{1}{( x - 1 ) ^{2}})

\int_{0}^{5} 4 e^{2 x} d x

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

What is the integral of 6/((x^2+1)^{3/2)} ?
The integral of 6/((x^2+1)^{3/2)} is (6x)/((1+x^2)^{1/2)}+C