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What happened to u all.please integrate it.....

i think i have done this

integrating by parts

I= root (1+x^3) int. 1dx - int. [d/dx(root(1+x^3).int. 1.dx]

I=x.root (1+x^3) - 3/2int. [x^3/root(1+x^3)]

take 1+x^3=t^2

3x^2dx=2t  dt

dx=2t/3x^2 dt

putting dx

I=x.root(1+x^3) - int. xdt

I=x.root(1+x^3) - int.(t^2-1)^1/3 dt

can anyone do after this.

 

only god can integrate it i think...

so put (1+x^3)=t^2

thus u get x=(t^2-1)^1/3

thus dx=(1/3)*(t^2-1)^(1/3 - 1) * 2t dt

thus dx=(2/3)*t/(t^2-1)^2/3

nw put t^2-1 =z

thus 2tdt=dz

thus ur integral reduces to ....

u can easily integrate it ................

Good work deedee ,  a nice approach

but when u assume 1+x^3=t^2 its root will be t ,where is it,,,,,,,,

yes t is missing, which makes your answer perfectly wrong.

SILLy me !!

(1+x^3)^1/2

so nw put (1+x^3)=t^2

so u get x=(t^2-1)^1/3

dx=(2/3)*t/(t^2-1)^(2/3)

so nw ur integral bcums ,.........................

so nw  put (t^2-1) =z so ur integral bcums (1/3)*(z-1)^1/2  / (z)^2/3

thinkin after this ........solly 4 dat wrng sol silly me bhul gayi

One of my teachers told it can't be integrated.its integration can't be found.........

So, i am telling the person who posted the question.........

Please don't ask such question..this is wasting others time as well as ur time.........

how did deedee post that soln

its not integrable i thnk...

may i ask y is it not integrable>??

Where deedee has posted the ful solution of that question.............