How do you integrate #int x/(sqrt(2x-1))dx# from [1,5]?

Question

How do you  integrate #int x/(sqrt(2x-1))dx# from [1,5]?

Ezekiel Alexander · Accepted Answer

The answer is #=16/3#

#x=1/2(2x-1)+1/2#

Let's rewrite the function

#x/(sqrt(2x-1))=(1/2(2x-1)+1/2)/(sqrt(2x-1))#

#=1/2*(2x-1)/(sqrt(2x-1))+1/(2sqrt(2x-1))#

#=1/2*sqrt(2x-1)+1/2*1/sqrt(2x-1)#

#int_1^5(xdx)/(sqrt(2x-1))=int_1^5 1/2*sqrt(2x-1)dx+int_1^5 1/2*dx/sqrt(2x-1)#

#= [ 1/2*(2x-1)^(3/2)/(2*3/2)+1/2*(2x-1)^(1/2)/(2*1/2) ]_1^5 #

#= [(2x-1)^(3/2)/6+(2x-1)^(1/2)/2 ]_1^5 #

#=(9/2+3/2)-(1/6+1/2)#

#=6-2/3=16/3#

Paisley Johnson · Answer

undefined To integrate the function $ \frac{x}{\sqrt{2x-1}} $ from 1 to 5, you can use the substitution method. Let $ u = \sqrt{2x-1} $, then $ x = \frac{u^2 + 1}{2} $ and $ dx = \frac{u}{\sqrt{2x-1}} \, du $.

Now, substitute these expressions into the integral:

$$ \int_{1}^{5} \frac{x}{\sqrt{2x-1}} \, dx = \int_{\sqrt{2(1)-1}}^{\sqrt{2(5)-1}} \frac{\frac{u^2+1}{2}}{u} \cdot \frac{u}{\sqrt{2x-1}} \, du $$

$$ = \int_{\sqrt{1}}^{\sqrt{9}} \frac{u^2+1}{2\sqrt{2u^2-1}} \, du $$

$$ = \int_{1}^{3} \frac{u^2+1}{2\sqrt{2u^2-1}} \, du $$

This integral can be further simplified using algebraic manipulations and trigonometric substitution. Once you find the antiderivative, evaluate it at the upper and lower limits of integration (3 and 1, respectively) and subtract the lower limit result from the upper limit result to find the final answer.