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April 18, 2014

April 18, 2014

Posted by **Ashley** on Monday, May 2, 2011 at 4:57am.

x/(2x^2 − 1)^0.4 dx

- Brief Calculus -
**drwls**, Monday, May 2, 2011 at 6:33amLet 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.

- Brief Calculus -
**Anonymous**, Monday, May 2, 2011 at 6:48am2*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

- Brief Calculus -
**Ashley**, Tuesday, May 3, 2011 at 3:24pmThanks Guys!

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