What is the integral of 2/(16+3x^2) ?

The integral of 2/(16+3x^2) is 1/(2sqrt(3))arctan((sqrt(3))/4 x)+C

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integral of 2/(16+3x^2)

\int \frac{2}{16 + 3 x ^{2}} d x

\frac{1}{2 3} arctan (\frac{3}{4} x) + C

Solution steps

\int \frac{2}{16 + 3 x ^{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}{16 + 3 x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \frac{1}{4 3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 \cdot \frac{1}{4 3} arctan (u)

Substitute back

u = \frac{3}{4} x

= 2 \cdot \frac{1}{4 3} arctan (\frac{3}{4} x)

Simplify

2 \cdot \frac{1}{4 3} arctan (\frac{3}{4} x) : \frac{1}{2 3} arctan (\frac{3}{4} x)

= \frac{1}{2 3} arctan (\frac{3}{4} x)

Add a constant to the solution

= \frac{1}{2 3} arctan (\frac{3}{4} x) + C

re g a r d s p \frac{I}{2 p}

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

\frac{d ^{2} y}{d x ^{2}} = 4 x

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

ln (x - 4)

What is the integral of 2/(16+3x^2) ?
The integral of 2/(16+3x^2) is 1/(2sqrt(3))arctan((sqrt(3))/4 x)+C