Recent content by irok

Evaluate the integral: \int \frac {-2x^{2} - 9x - 50}{x^{3} + 8x^{2} + 30x + 36} x^{3} + 8x^{2} + 30x + 36 = (x+2)(x^{2}+6x+18) \int \frac {-2x^{2} - 9x - 50}{(x+2)(x^{2}+6x+18)} I'm stuck at this point. I tried finding roots for x^{2}+6x+18 but can't. tried dividing x^{2}+6x+18 with...

Evaluate the Integral: \int \frac {2x+1}{(x^{2}+9)^{2}} My attempt: \frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}} = (Ax+B)(x^{2} + 9)^{2} + (Cx+D)(x^{2} + 9) = Ax^{5} + Bx^{4} Dx^{3} + (18A + E)x^{2} + (81A+9D+18B)x + 9E + 81B I'm not sure what...

I ended up with x(ln3x)^{2} - 2\int ln(3x) u = (ln(3x))^{2} du = 2 ln(3x) \frac {1}{x} Not sure how to integrate ln(3x). I know integral of ln(x) is xln(x)-x+C. Still not sure what to do when it's ln(3x). I am now assuming any integration of ln(kx) is 1/x

#2. \int (ln(3x))^{2} dx I'm still stuck on this one. Do I use \int ln(3x) ln(3x) and f=ln(3x) g'=ln(3x) or \int (ln(3x))^{2} and f=ln(3x)^{2} g'=1

Cool, at first I didn't understand why it didn't matter if ln(2x) or ln(kx). Now I'm all cleared up! Thank you again rootX

#5. Use integration by parts to evaluate the definite integral. \int t e^{-t} dt Is the following correct? f = t f' = 1 g' = e^{-t} g = -e^{-t} -t e^{-t} ]^{1}_{0} - [e^{-t}]^{1}_{0} Solved Thanks Rocomath.

#1. Not sure how that way works. But can you confirm if the following are correct: f=2x f'=2 g'=ln(2x) g=1/(2x) #2. The anti-derivative for ln(3x) = 1/3x? f=ln(3x) f'=(1/x) g'=ln(3x) g=? #3. SOLVED I did something wrong when first trying this question. I didn't integrate e^4x properly so i...

Homework Statement #1. Use Integration by parts to evaluate the integral \int 2x \ln(2x) dx #2. Use Integration by parts to evaluate the integral \int (\ln(3x))^{2} dx #3. Use Integration by parts to evaluate the integral \int x e^{4x} dx #4. Evaluate the indefinite integral. \int \sin(3x)...

Yep, I got Question #1. I made u = sinx+3 instead of u = sinx. Thank you rocomath! One more question, can i simplify ln(ln(x))?

\int \frac{2 \; dx}{x \ln (6 x)} Let u = ln(6x), du = 1 / 6x = 12 \int \frac{1}{u} \ du = 12 [\ln(u)] = 12 [\ln(ln(6x))] + C Are there any mistakes?

Homework Statement Question 1: Evaluate the indefinite integral. \int \frac{\cos x}{2 \sin x + 6} \, dx Question 2: Evaluate the indefinite integral. \int \frac{2 \; dx}{x \ln (6 x)} NOTE: The absolute value of x has to be entered as abs(x). The Attempt at a Solution Question 1...

Well, for Question one: Can anyone confirm that f(x) = 6x^{5} and a = 2. I'm pretty sure that a = 2 since, F(x) - F(a) = [ 6x^{5} / x^{6} ] - [ 2 ] = 6x^{-1} - 2

Homework Statement Question One: Find a continuous function f and a number a such that 2 + \int_{a}^{x} \frac {f(t)} {t^{6}} \,dt = 6 x^{-1} Question Two: At what value of x does the local max of f(x) occur? f(x) = \int_0^x \frac{ t^2 - 25 }{ 1+\cos^2(t)} dt The attempt at a solution I...