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

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

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

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

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

Solution steps

\int \frac{2}{( 2 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

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

Apply Trigonometric Substitution

: \int \frac{1}{2 ^{\frac{1}{2}} sec ( u )} d u

= 2 \cdot \int \frac{1}{2 ^{\frac{1}{2}} sec ( u )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2 ^{\frac{1}{2}}} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= 2 \cdot \frac{1}{2 ^{\frac{1}{2}}} \cdot \int cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{2 ^{\frac{1}{2}}} sin (u)

Substitute back

u = arctan (2^{\frac{1}{2}} x)

= 2 \cdot \frac{1}{2 ^{\frac{1}{2}}} sin (arctan (2^{\frac{1}{2}} x))

Simplify

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

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

Add a constant to the solution

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

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