How do you integrate $\int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}}$ $int 1/sqrt(2x-12sqrtx-5)$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}}$ $int 1/sqrt(2x-12sqrtx-5)$ using trigonometric substitution?

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Answer 1 · 2018-05-02T12:54:56

$I = \sqrt{2 x - 12 \sqrt{x} - 5} + 3 \sqrt{2} ln ∣ ∣ \sqrt{2 x} - 3 \sqrt{2} + \sqrt{2 x - 12 \sqrt{x} - 5} ∣ ∣ + C,$ $I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,$
There is no any trig.function in the answer .So it is difficult to proceed with trig substitution.

Explanation:

Here,

$I = \int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}} d x$ $I=int1/sqrt(2x-12sqrtx-5)dx$

Let, $\sqrt{x} = t \Rightarrow x = t^{2} \Rightarrow d x = 2 t d t$ $sqrtx=t=>x=t^2=>dx=2tdt$

$I = \int \frac{2 t}{\sqrt{2 t^{2} - 12 t - 5}} d t$ $I=int(2t)/sqrt(2t^2-12t-5)dt$

$= \frac{1}{2} \int \frac{4 t - 12 + 12}{\sqrt{2 t^{2} - 12 t - 5}} d t$ $=1/2int(4t-12+12)/sqrt(2t^2-12t-5)dt$

$= \frac{1}{2} \int \frac{4 t - 12}{\sqrt{2 t^{2} - 12 t - 5}} d t + \frac{1}{2} \int \frac{12}{\sqrt{2 t^{2} - 12 t - 5}} d t$ $=color(red)(1/2int(4t-12)/sqrt(2t^2-12t-5)dt)+color(blue) (1/2int12/sqrt(2t^2-12t-5)dt$

$I=color(red)(I_1)+color(blue)(I_2)...to(A)$

Now, $I_1=1/2int(4t-12)/sqrt(2t^2-12t-5)dt$

Take, $sqrt(2t^2-12t-5)=u=>2t^2-12t-5=u^2$

$=>4t-12=2udu$

$:.I_1=1/2int(2u)/udu=intdu=u+c_1,tou=sqrt(2t^2-12t-5)$

$=sqrt(2t^2-12t-5)+c_1,where,sqrtx=t$

$:.I_1=sqrt(2x-12sqrtx-5)+c_1$

$Also, I_2=1/2int12/sqrt(2t^2-12t-5)dt$

$I_2=6/sqrt2int1/sqrt(t^2-6t-5/2)dt$

$=6/sqrt2int1/sqrt(t^2-6t+9-23/2)dtto[-5/2=(18-23)/2]$

$=3sqrt2int1/sqrt((t-3)^2-(sqrt(23/2))^2)dt$

$=3sqrt2ln|t-3+sqrt((t-3)^2-(sqrt(23/2))^2)|+c$

$=3sqrt2ln|t-3+color(violet)(sqrt(t^2-6t-5/2))|+c$

$=3sqrt2ln|t-3+sqrt(2t^2-12t-5)/color(orange)(sqrt2)|+color(orange)(c$

$I_2=3sqrt2ln|sqrt2t-3sqrt2+sqrt(2t^2-12t-5)|+color(orange)(c_2$

Subst .back $t=sqrtx$

$I_2=3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx-5)|+c_2$

Hence, from $(A)$ ,

$I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,$

$where,C=c_1+c_2$

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How do you integrate $\int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}}$ $int 1/sqrt(2x-12sqrtx-5)$ using trigonometric substitution?

1 Answer

Explanation:

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How do you integrate int 1/sqrt(2x-12sqrtx-5) ∫1√2x−12√x−5int 1/sqrt(2x-12sqrtx-5) using trigonometric substitution?

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Explanation:

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How do you integrate $\int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}}$ $int 1/sqrt(2x-12sqrtx-5)$ using trigonometric substitution?