Brief Calculus
posted by Ashley
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

drwls
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. 
Anonymous
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(t1)^0.4=
(1/4) Integral of (t1)^(0.4)dt
Integral of x^n=x^(n+1)/(n+1)
Integral of (t1)^(0.4)dt=
(t1)^(0.4+1)/(0.4+1)+C=
(t1)^0.6/0.6+C=(t1)^(3/5)/0.6+C
Integral of x/(2x^2−1)^0.4 dx=
(1/4) Integral of (t1)^(0.4)dt=
(1/4)(t1)^(3/5)/0.6+C=
(1/4*0.6)(t1)^(3/5)=
(1/2.4)(t1)^(3/5)+C=
0.41666666(t1)^(3/5)+C
t=2x^2
0.41666666(t1)^(3/5)+C=
0.4167(2x^21)^(3/5)+C rounded to 4 decimal pieces
Integral of x/(2x^2−1)^0.4 dx=
0.4167(2x^21)^(3/5)+C 
Ashley
Thanks Guys!
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