# Calculus Integration, using the substitution method

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Well, have you tried a substitution? – user61527 Oct 18 '13 at 7:49
Yes, multiple times. However I cannot quite pick the right function to represent 'u' in order to cancel out the 'x' values. – James Oct 18 '13 at 7:50
$2xdx=dx^{2}$ bells ringing? – drhab Oct 18 '13 at 8:05
No bells ringing, have just started integration, the only tools I have to solve are the reverse of the power rule and u-substitution. Wolfram Alpha has gone off into inserting trig identities, what I am looking for is a framework I can apply to approach this problem and future ones... At this stage I get to the point where 8∫x(x^2-1)(2x^2-3)^-(1/2)dx... Thus far I have been able to get by looking for the derivative of one of the functions and cancelling out the x values, however I have been unable to successfully do so this time. – James Oct 18 '13 at 8:15

Make the change : $2 x^2 - 3 = y^2$. The integral becomes very simple.

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Thank you very much, thank was exactly the hint needed - it has helped a lot. My final answer keeps coming up as 4/3(2x^2-3)+(2x^2-3)^(1/2)+C however the back of the book advises 4/3*x^2(2x^2-3)^(1/2)+C. I will continue working through it tomorrow. Thanks again – James Oct 18 '13 at 9:21
@James Assuming that $2x^2-3$ is not in the denominator, you could clear the parentheses and add to C to form a new constant. Both answers would be equivalent. – Mike Oct 18 '13 at 9:36

If you make the change 2x^2-3=y^2, you have to integrate (2+2y^2)dy which is effectively very simpple. After the integration, replace y by its expression as a function of x and continue simplifying. You then arrive to 4x^2 Sqrt[2x^2-3]/3; this is what tells your book . Add the integration constant to make your book happy ! Simple, isn't ? I am glad to have been able to help you.

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