What is the integral of 6/(x^2+4) ?

The integral of 6/(x^2+4) is 3arctan(x/2)+C

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integral of 6/(x^2+4)

\int \frac{6}{x ^{2} + 4} d x

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

Solution steps

\int \frac{6}{x ^{2} + 4} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 6 \cdot \int \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= 6 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

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

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

Add a constant to the solution

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

x \to 1 lim (\frac{x ^{2} + 1}{x - 1})

x \to 3 lim (2 ln (x - 4))

\int 12 (y - 1) x^{3} e^{2 x (y - 1)} d y

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

f (x) = (x^{2} - 3) (x^{2} - 2)