What is the integral of e^4x*tan^2(x) ?

The integral of e^4x*tan^2(x) is e^4(-x^2+xtan(x)+(x^2)/2+ln|cos(x)|)+C

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integral of e^4x*tan^2(x)

integral

e^{4} x \cdot tan^{2} (x)

e^{4} (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣) + C

Solution steps

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

Take the constant out:

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

= e^{4} \cdot \int x tan^{2} (x) d x

Apply Integration By Parts

: x (- x + tan (x)) - \int - x + tan (x) d x

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

\int - x + tan (x) d x = - \frac{x ^{2}}{2} - ln ∣ cos (x) ∣

= e^{4} (x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣))

Expand

x (- x + tan (x)) - (- \frac{x ^{2}}{2} - ln ∣ cos (x) ∣) : - x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣

= e^{4} (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣)

Add a constant to the solution

= e^{4} (- x^{2} + x tan (x) + \frac{x ^{2}}{2} + ln ∣ cos (x) ∣) + C

y = \frac{x}{2 + 1}

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

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

\frac{1 + x}{1 - x}

(1 - x)^{5}

What is the integral of e^4x*tan^2(x) ?
The integral of e^4x*tan^2(x) is e^4(-x^2+xtan(x)+(x^2)/2+ln|cos(x)|)+C