What is the integral of 1/(7+4x^2) ?

The integral of 1/(7+4x^2) is 1/(2sqrt(7))arctan(2/(sqrt(7))x)+C

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integral of 1/(7+4x^2)

\int \frac{1}{7 + 4 x ^{2}} d x

\frac{1}{2 7} arctan (\frac{2}{7} x) + C

Solution steps

\int \frac{1}{7 + 4 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{2 7 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2 7} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{2 7} arctan (u)

Substitute back

u = \frac{2}{7} x

= \frac{1}{2 7} arctan (\frac{2}{7} x)

Add a constant to the solution

= \frac{1}{2 7} arctan (\frac{2}{7} x) + C

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What is the integral of 1/(7+4x^2) ?
The integral of 1/(7+4x^2) is 1/(2sqrt(7))arctan(2/(sqrt(7))x)+C