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.#