Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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


4 Answers

Khushan Sanghvi
29 Points
2 years 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)
Sai Ram Charan
31 Points
2 years ago
Hello Vinod! did you mean the following integral?
\int \frac{1}{\sqrt{x(x^6-16)}}dx
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.
Sai Ram Charan
31 Points
2 years 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.
Samyak Jain
333 Points
2 years ago
Let T = (1/x\sqrt{x^6-16} dx = ∫x2 / x3\sqrt{x^6-16} 
Let x3 = 1/t  \Rightarrow 3x2dx = – 1/t2dt  i.e. x2dx = – 1/3t2dt 
So, T = [(–1/3t2)/{(1 / t)\sqrt{1/t^2 - 16}}] dt = (–1/3) 1 /[ t\sqrt{1 - 16t^2} / t]
         = (–1/3) [1 /\sqrt{1 - 16t^2}] dt = (1/3) [–1 /\sqrt{1 - (4t)^2}] dt
T = (1/3)(1/4)cos–1(4t) + c  =  (1/12)cos–1(4t) + c
[\because [–1 /\sqrt{1 - (x)^2}] dx = cos–1x + c]

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy


Get your questions answered by the expert for free