How do you integrate #int 1/sqrt(4x^2+4x+-24)dx# using trigonometric substitution?

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Answer 1 · 2016-06-28T13:18:14

#I=ln|(2x+1+sqrt(4x^2+4x-24))/5|+C,#

OR

#I=ln|2x+1+sqrt(4x^2+4x-24)|+K,#
where, #K=C-ln5.#

I will solve #I=int1/sqrt(4x^2+4x-24)dx.#

We have, #4x^2+4x-24=4x^2+4x+1-25=(2x+1)^2-5^2.#

Knowing that, #sec^2theta-1=tan^2theta, 2x+1=5sect# will be proper substitution.

So, let, #2x+1=5sect............(1),# so that, #2dx=5*sect*tantdt,# and,

#sqrt(4x^2+4x-24)=sqrt{(2x+1)^2-5^2}=sqrt(25sec^2t-25)=5tant...........(2).#

Hence, #I=int1/sqrt(4x^2+4x-24)dx=int1/(5tant)*5*sect*tantdt=intsectdt=ln|sect+tant|.#

Hence, by #(1) & (2), I=ln|(2x+1+sqrt(4x^2+4x-24))/5|+C.#

One may say #I=ln|2x+1+sqrt(4x^2+4x-24)|+K,#
where, #K=C-ln5.#