How do you find the integral of $int 4/(1+9x^2)dx$ ?

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Answer 1 · 2018-03-18T20:27:51

$4/3tan^(-1)(3x)+c$

$int4/(1+9x^2)dx$

substitute

$9x^2=tan^2u$

$=>3x=tanu--(1)$

$=>3dx=sec^2udu$

$int4/(1+9x^2)dx=4int1/(1+tan^2u)xx(sec^2udu)/3$

$=4/3intcancel((sec^2u)/(1+tan^2u))du$

$I=4/3intdu=4/3u+c$

$(1)rarru=tan^(-1)(3x)$

$:.I=4/3tan^(-1)(3x)+c$

How do you find the integral of int 4/(1+9x^2)dxint 4/(1+9x^2)dx?