What is the integral of 5tan^2(x)+tan^4(x) ?

The integral of 5tan^2(x)+tan^4(x) is -4x+1/3 tan^3(x)+4tan(x)+C

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integral of 5tan^2(x)+tan^4(x)

\int 5 tan^{2} (x) + tan^{4} (x) d x

- 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x) + C

Solution steps

\int 5 tan^{2} (x) + tan^{4} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 5 tan^{2} (x) d x + \int tan^{4} (x) d x

\int 5 tan^{2} (x) d x = 5 (- x + tan (x))

\int tan^{4} (x) d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 5 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x)

Simplify

5 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x) : - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x)

= - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x)

Add a constant to the solution

= - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x) + C

\int e^{(- x^{2})} d x

y = x^{2}, y = 2 - x, x = 0

f (x) = ln (\frac{e ^{x}}{1 + e ^{x}})

y = ln (\frac{3}{x})

n = 0 \sum \infty \frac{2}{n + 1}

What is the integral of 5tan^2(x)+tan^4(x) ?
The integral of 5tan^2(x)+tan^4(x) is -4x+1/3 tan^3(x)+4tan(x)+C