What is the integral of #int ((x^2-1)/sqrt(2x-1) )dx #?

Question

Emma Aldridge · Accepted Answer

#int\ (x^2-1)/sqrt(2x-1)\ dx=1/20(2x-1)^(5/2)+1/6(2x-1)^(3/2)-3/4sqrt(2x-1)+C# Our big problem in this integral is the root, so we want to get rid of it. We can do this by introducing a substitution #u=sqrt(2x-1)#. The derivative is then #(du)/dx=1/sqrt(2x-1)# So we divide through (and remember, dividing by a reciprocal is the same as multiplying by just the denominator) to integrate with respect to #u#: #int\ (x^2-1)/sqrt(2x-1)\ dx=int\ (x^2-1)/cancel(sqrt(2x-1))cancel(sqrt(2x-1))\ du=int\ x^2-1\ du# Now all we need to do is express the #x^2# in terms of #u# (since you can't integrate #x# with respect to #u#): #u=sqrt(2x-1)# #u^2=2x-1# #u^2+1=2x# #(u^2+1)/2=x# #x^2=((u^2+1)/2)^2=(u^2+1)^2/4=(u^4+2u^2+1)/4# We can plug this back into our integral to get: #int\ (u^4+2u^2+1)/4-1\ du# This can be evaluated using the reverse power rule: #1/4*u^5/5+2/4*u^3/3+u/4-u+C# Resubstituting for #u=sqrt(2x-1)#, we get: #1/20(2x-1)^(5/2)+1/6(2x-1)^(3/2)-3/4sqrt(2x-1)+C#

Ezra Adams · Answer

undefined To find the integral of $\int \frac{x^2-1}{\sqrt{2x-1}} \, dx$, we can use the method of substitution. Let $u = 2x-1$, then $du = 2 \, dx$. Rearranging for $dx$, we get $dx = \frac{1}{2} du$. Substituting these into the integral, we have:

$\int \frac{x^2-1}{\sqrt{2x-1}} \, dx = \int \frac{x^2-1}{\sqrt{u}} \cdot \frac{1}{2} \, du$

Expanding $x^2-1$ and factoring out $\frac{1}{2}$, we obtain:

$\frac{1}{2} \int \left(\frac{x^2}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) \, du$

Splitting the integral into two separate integrals, we get:

$\frac{1}{2} \left(\int \frac{x^2}{\sqrt{u}} \, du - \int \frac{1}{\sqrt{u}} \, du\right)$

Now, we integrate each term separately:

For $\int \frac{x^2}{\sqrt{u}} \, du$, we can rewrite $x^2$ in terms of $u$. Since $u = 2x-1$, we have $x^2 = \frac{u+1}{2}$. Substituting this, we get:

$\int \frac{\frac{u+1}{2}}{\sqrt{u}} \, du = \frac{1}{2} \int \frac{u+1}{\sqrt{u}} \, du$

This integral can be solved using u-substitution.

For $\int \frac{1}{\sqrt{u}} \, du$, we have a simple integral:

$\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C$

Substituting back $u = 2x-1$, we get the final result:

$\frac{1}{2} \left(\int \frac{x^2}{\sqrt{2x-1}} \, dx - 2\sqrt{2x-1}\right) + C$

This is the integral of $\int \frac{x^2-1}{\sqrt{2x-1}} \, dx$.