Calculus

posted by .

I have to use integration by parts to integrate tan^-1 (1/x)dx. I'm not sure what to use as u or dv. We were taught to use uv-integral of (vdu) to do this.

  • Calculus -

    look for a u that does not get too messy on differentiation, and a v that can be integrated without too much troubls

    here, it seems logical to use u = arctan(1/x) since we know we're going to get rid of the nasty arctan, and the chain rule will toss in a 1/x^2:

    u = arctan(1/x)
    du = 1/(1/x^2 + 1)* (-1/x^2) dx
    = -1/(1+x^2) dx

    dv = dx
    v = x

    uv-Int v*du = x*arctan(1/x) + Int(x/(1+x^2)) dx

    Int v*du = 1/2 * ln(1+x^2)

    So, the final answer is x*arctan(1/x) + 1/2 ln(1+x^2)

    go ahead -- take the derivative and watch things cancel out!!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus - Integration

    Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I …
  2. calculus-integration!

    should i use substitution?? if yes how should should i use it?
  3. calculus-integration

    integrate -2/xln^4(x)...plz help me..give me an idea on how to start..plz The derivative of the ln(x) function is 1/x and this is multiplying the ln^4(x). You can thus write the integral as: -2 * 1/5 ln^5(x) + constant. is that the …
  4. calc

    evaluate the integral: y lny dy i know it's integration by parts but i get confused once you have to do it the second time Leibnitz rule (a.k.a. product rule): d(fg) = f dg + g df y lny dy = d[y^2/2 ln(y)] - y/2 dy ----> Integral …
  5. calc

    also: integral of tan^(-1)y dy how is integration of parts used in that?
  6. Math/Calculus

    How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this method. …
  7. calculus

    Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)-[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3 - integral of(1/3)tan(3x)dx - (1/3)[ln(sec(3x))/3] …
  8. Calculus

    "Evaluate the following indefinite integral using integration by parts: *integral sign* tan^-1(x) dx" I let u = tan^-1(x) and dv = dx. Is that right?
  9. Calculus II

    Integrate using integration by parts (integral) (5-x) e^3x u = 5-x du = -dx dv = e^3x v = 3e^3x I wonder if this is right so far. = uv - (integral) v du = (5-x)(3e^3x) - (integral) (-3e^3x) =(5-x)(3e^3x) + (integral) (3e^3x) = (5-x)(3e^3x) …
  10. Calculus

    Use integration by parts to evaluate the integral xsqrt(2x+6)dx.

More Similar Questions