$$\int{(24ab-c-dx)^{3/2}\over a^2-b^2x^2} dx $$ I have tried to solve this integral for a case when $$\ dx=constant$$ inside in a bracket but i am stuck that how to deal this generally.Thanks in advanced.
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2What do you mean with $\ dx=constant$ ? – Claude Leibovici Sep 25 '17 at 07:40
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mean if this whole is constant (24ab−c−constant)3/2 then it will be solve other wise i am stuck please give me hints – umairkhan Sep 25 '17 at 07:45
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2OK, I see now : you mean $d \times x=constant$ not $dx=constant$. These notations are confusing. – Claude Leibovici Sep 25 '17 at 07:48
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yes sir d is constant mean d.x .simply d multiply with x thanks sir – umairkhan Sep 25 '17 at 07:58
1 Answers
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Hint:)
Let $\dfrac{2ab-c}{d}=A$, $\dfrac{a}{b}=B$ and then substituation $A-x=u^2$ $$ \int{(24ab-c-dx)^{3/2}\over a^2-b^2x^2} dx = \dfrac{d^\frac32}{b^2}\int\dfrac{(A-x)^\frac32}{B^2-x^2}dx = -2\dfrac{d^\frac32}{b^2}\int\dfrac{u^4}{(B+A-u^2)(B-A+u^2)}du $$
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