What is the integral of e^xtan^4(e^x) ?

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

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integral of e^xtan^4(e^x)

\int e^{x} tan^{4} (e^{x}) d x

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

Solution steps

\int e^{x} tan^{4} (e^{x}) d x

Apply u-substitution

: \int tan^{4} (u) d u

= \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u

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

= \frac{tan ^{3} ( u )}{3} - (- u + tan (u))

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{6}{( 2 - x ) ^{4}})

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

\frac{\partial}{\partial y} (x^{2} - e^{y^{2}})

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

\int - t^{3} d t

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