What is the integral of 4/((x^2+1)^{3/2)} ?

The integral of 4/((x^2+1)^{3/2)} is (4x)/((1+x^2)^{1/2)}+C

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integral of 4/((x^2+1)^{3/2)}

\int \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{4 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{4}{( 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

= 4 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= 4 \cdot \int cos (u) d u

Use the common integral:

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

= 4 sin (u)

Substitute back

u = arctan (x)

= 4 sin (arctan (x))

Simplify

4 sin (arctan (x)) : \frac{4 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{4 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{4 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (\frac{x ^{5}}{x ^{3}})

x \to 9 lim ([f (18) - g (7)])

n = 2 \sum \infty ln (\frac{1}{2}) ln (n)

\frac{e ^{x} + e ^{- x}}{2}

x \to 0 + lim (e^{2 + l n (x)})

What is the integral of 4/((x^2+1)^{3/2)} ?
The integral of 4/((x^2+1)^{3/2)} is (4x)/((1+x^2)^{1/2)}+C