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Use the product rule: d/dx(uv) = uv' + u'v, with u = x^2 and v = e^(tan(x)), so that u' = 2x and v' = sec^2(x) * e^(tan(x)), and so the answer is 2x * e^(tan(x)) + x^2 * sec^2(x) * e^(tan(x)) .

First step: partial fractions 5/(2x-1)(x-3) 5=A(x-3)+B(2x-1) A=0 when x=3, so B=5/(2x3-1)=1 B=0 when x=1/2, so A=5/(0.5-3)=-2 So f'(x)=1/(x-3)-2/(2x-1) Second step: Integration f(x)= (integral)(1/(x-3))dx...

you need to think about how the integral will be simpliefied, basically you are trying to get to a state where only one function of x resides in the integral. so if you have an x² sin3x dx insi...

y′=3x2 −6x

use of y′ = 0

0= 3x^2 - 6x

0= 3x(x-2)

therefore either x=0 or x=2

when x=0 y=1, when x=2 y=-3

(0, 1) or (2, −3) 

y''=6x-6

when x=0 y''=6(...

Start with the LHS:

(cos^4x - sin^4x) / cos^2x 

Recognise the difference of two squares on the top line, which simplifies to (cos^2x - sin^2x)(cos^2x + sin^2x):

(cos^2x - sin^2x)(cos^...