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
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.
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
Thanks Guys!
To evaluate the given integral, we can use the substitution method. Let's begin by selecting an appropriate substitution.
In this case, a suitable substitution would be to let u = 2x^2 - 1. This substitution allows us to express the expression within the integral in terms of u, which will make it easier to integrate.
Let's find the derivative of u with respect to x:
du/dx = 4x
Next, we can solve for dx:
dx = du/(4x)
Now, we can substitute the values of u and dx into the integral:
∫ (x/(2x^2 - 1)^0.4) dx
Becomes:
∫ (x/(u^0.4)) * (du/(4x))
Simplifying the expression:
∫ (1/(4u^0.4)) du
We can now integrate ∫ (1/(4u^0.4)) du. The integral of u^(-n) with respect to u is (u^(-n+1))/(1-n), so substituting n = 0.4, we have:
(1/(1-0.4)) * (∫ (u^(-0.4)) du)
Simplifying further:
(1/0.6) * (u^(0.6))/(0.6)
Simplifying again:
(5/3) * (u^(0.6))
Now, we substitute back for u:
(5/3) * ((2x^2 - 1)^(0.6))
Finally, we have evaluated the integral using the substitution method.