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        integrate ∫(1/(x(x^6-16)^1/2))dx pls  solve it and send the solution
11 months ago

							∫1/(x(x^6-16)^1/2)) dx     Multiply and divide by x^2  u get  ∫x^2 / x^3((x^6-16)^1/2)     Put x^3=t     dt/3=x^2dx   So now ur integration is∫1/t((t^2-16)^1/2))dt    Take the t outside the root inside u get  ∫1/((t^4-16t^2)^1/2))   Now complete the square and u will get        ∫1/(((t^2-8)^2-8^2)^1/2) now using general formula                                                  ∫1/(x^2-a^2)^1/2 dx = log(x+(x^2-a^2)^1-2)            Here x is t^2-8 and a is 8    So substitute and substitute back value of x^3 which is t.  Ur final ans will be                                    log(x^6-8+(((x^6-8)^2)-8^2)^1/2)

10 months ago
							Hello Vinod! did you mean the following integral?For this integral there exists no closed function as an answer- i.e., the integral isn’t elementary function!This is actually out of class 12 syllabus I think.

10 months ago
							We can answer it if it has limits. Many indefinite integrals donot have closed answers.No antiderivative could be found without limit, or all supported integration methods were tried unsuccessfully. Note that many functions don't have an elementary antiderivative. Antiderivative or integral could not be found. Note that many functions don't have an elementary antiderivative. I’m telling again, we can answer this qeustion if it has limits.

10 months ago
							Let T = ∫(1/x dx = ∫x2 / x3 Let x3 = 1/t   3x2dx = – 1/t2dt  i.e. x2dx = – 1/3t2dt So, T = ∫[(–1/3t2)/{(1 / t)}] dt = (–1/3)∫ 1 /[ t / t]         = (–1/3)∫ [1 /] dt = (1/3)∫ [–1 /] dtT = (1/3)(1/4)cos–1(4t) + c  =  (1/12)cos–1(4t) + c[ ∫[–1 /] dx = cos–1x + c]

9 months ago
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