Brief Calculus

Decide on what substitution to use, and then evaluate the given integral using a substitution. HINT [See Example 1.] (Round your decimal coefficients to four decimal places.)
x/(2x^2 − 1)^0.4 dx


asked by Ashley
  1. Let u = 2x^2 -1
    Then du = 4x dx, and x*dx = du/4,

    x/(2x^2 − 1)^0.4 dx = (du/4)*u^-0.4

    You can easily integrate that.

    posted by drwls
  2. 2*x^2=t

    2*2xdx=dt

    4xdx=dt Divide with 4

    xdx=dt/4

    Integral of x/(2x^2−1)^0.4 dx=

    Integral of dt/4(t-1)^0.4=

    (1/4) Integral of (t-1)^(-0.4)dt

    Integral of x^n=x^(n+1)/(n+1)

    Integral of (t-1)^(-0.4)dt=

    (t-1)^(-0.4+1)/(-0.4+1)+C=

    (t-1)^0.6/0.6+C=(t-1)^(3/5)/0.6+C


    Integral of x/(2x^2−1)^0.4 dx=

    (1/4) Integral of (t-1)^(-0.4)dt=

    (1/4)(t-1)^(3/5)/0.6+C=

    (1/4*0.6)(t-1)^(3/5)=

    (1/2.4)(t-1)^(3/5)+C=

    0.41666666(t-1)^(3/5)+C

    t=2x^2

    0.41666666(t-1)^(3/5)+C=

    0.4167(2x^2-1)^(3/5)+C rounded to 4 decimal pieces


    Integral of x/(2x^2−1)^0.4 dx=

    0.4167(2x^2-1)^(3/5)+C

    posted by Anonymous
  3. Thanks Guys!

    posted by Ashley

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