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

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

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

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

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

Solution steps

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

Simplify

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{3}{2} x

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

Add a constant to the solution

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

f (x) = 2 x^{3} - 24 x^{2} + 97 x + 7

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

x \to 10 lim (e^{- x})

\frac{d ^{2}}{d x ^{2}} (5 (x^{2} + 1)^{4})

\int_{2}^{5} - x^{2} + 7 x - 10 d x

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