Definite integral by change of variable

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[@Amoeby]

For a general case of the integral of (x^m)(a + bx^n)^p where m, n and p are rational numbers and a and b are real numbers: 1) if p is an integer then substitute x = t^(LCM of the denominators of m and n); 2) if (m + 1)/n is integer then substitute a + bx^n = t^q where q is the denominator of p; 3) if (m + 1)/n + p is an integer then substitute b + a/x^n = t^q where q is tge denominator of p. So after rewriting this integral in the binomial expression we get the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx. m = 3, n = 2, p = -3/2. 1) p is not an integer 2) (m + 1)/n = (3 + 1)/2 = 2 is an integer so 9 + 4x^2 = t^2 4xdx = tdt xdx = tdt/4 x^2 = (t^2 - 9)/4 x^3dx = t(t^2 - 9)dt/16 x = 0 -> t = 3, x = 3sqrt(3)/2 -> t = 6. Finally, the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx is equal to the integral from 3 to 6 of t(t^2 - 9)dt/(16t^3) = (1 - 9t^(-2))dt/16 = (t + 9/t)/16 where t goes from 3 to 6 so it equals to (6 + 9/6 - 3 - 9/3)/16 = 3/32.

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[@odumosuadeniyilukman]

Sir, I want to ask that condition will I find the value of the limits, or is it every time I have definite integrals, I will take their new value of u. Coz, I don't understand that. I just want to know the condition behind it. 🙇

[@AliHassan-hb1bn]

You could use integration by parts method

[@tah_aki]

Yeah.. better to know them both as the question can be direct sometimes 😊

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[@holyshit922]

We can factor from numerator derivative of the inside of denominator and we are left with x^2 whch can be expressed as function of sqrt(4x^2+9) so change of variable u=sqrt(4x^2+9) is good idea

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[@pgray380]

I get the same result without the x^3=x^2 * x trick. At one point I get [sqrt(u-9)]^3 in the numerator and sqrt(u-9) in the denominator. After cancellation left with [sqrt(u-9)]^2 which conveniently is just u-9.

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[@eliacoldwar-us9qr]

Find d²y/dx² when y=square root of x/(x²+x)

[@harshplayz31882]

Its just 1st step game Aftee u know u have break x³ to x² and x. After it its easy

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