Evaluate the integral with hyperbolic or trigonometric substitution. ?

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Answer 1 · 2018-04-26T16:42:50

$int1/(1+9x^2)dx=1/3arctan3x+"c"$

For $int1/(1+9x^2)dx=int1/(1+(3x)^2)dx$ , let

$u=3x$ and $du=3dx$

Then

$int1/(1+9x^2)dx=1/3int1/(1+u^2)du$

This is a standard integral which evaluates to

$1/3arctanu+"c"=1/3arctan3x+"c"$

Answer 2 · 2018-04-26T16:45:50

$1/3arc tan3x+C.$

Let, $I=int1/(1+9x^2)dx$ .

Knowing that $tan^2y+1=sec^2y$ , we subst. $3x=tany$ .

$:. 3dx=sec^2ydy$ .

$:. I=int1/(1+tan^2y)*sec^2y/3dy$ ,

$=1/3intsec^2y/sec^2ydy$ ,

$=1/3int1dy$ ,

$=1/3y$ .

$rArr I=1/3arc tan3x+C.$