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

The integral of (sec^2(x))/(6+2tan^2(x)) is 1/(2sqrt(3))arctan(1/(sqrt(3))tan(x))+C

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integral of (sec^2(x))/(6+2tan^2(x))

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

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}{1 + v ^{2}} d v

Use the common integral:

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

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

Substitute back

= \frac{1}{2 3} arctan (\frac{2}{6} tan (x))

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{1}{2} b (\frac{1}{x} - \frac{1}{x ^{0}})^{2}

f (x) = x^{3} - 5 x^{2} + 7, at x = 1

\int \frac{x + 5}{x ^{2} + 2 x - 15} d x

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

x \to 5 + lim (x^{2} - 25)

What is the integral of (sec^2(x))/(6+2tan^2(x)) ?
The integral of (sec^2(x))/(6+2tan^2(x)) is 1/(2sqrt(3))arctan(1/(sqrt(3))tan(x))+C