How do you integrate ? 1/(x^2+9)^(1/2)

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Answer 1 · 2018-05-27T10:54:31

$y=int1/sqrt(x^2+9) dx$

put $x=3 tant$ $rArr t=tan^-1(x/3)$
Hence, $dx= 3sec^2tdt$

$y=int(3sec^2t)/sqrt(9tan^2t+9) dt$

$y=int(sec^2t)/sqrt(tan^2t+1) dt$

$y=int(sec^2t)/sqrt(sec^2t) dt$

$y=int(sec^2t)/(sect) dt$

$y=int(sect) dt$

$y=ln|sec t + tan t| + C$

$y=ln|sec( tan^-1(x/3) )+ tan (tan^-1(x/3))| + C$

$y=ln|sec( tan^-1(x/3) )+ x/3)| + C$

$y=ln|sqrt(1+x^2/9 )+ x/3| + C$

Answer 2 · 2018-05-27T11:10:33

We know that,
$int1/sqrt(X^2+A^2) dX=ln|X+sqrt(X^2+A^2)|+c$
So,
$I=int1/(x^2+9)^(1/2) dx=int1/sqrt(x^2+3^2)dx$
$=>I=ln|x+sqrt(x^2+9)|+c$

$II^(nd)$ method : Trig. subst.

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

Take, $x=3tanu=>dx=3sec^2udu$

$and color(blue)(tanu=x/3$

So,

$I=int1/(9tan^2u+9)^(1/2) 3sec^2udu$

$=int(3sec^2u)/((9sec^2u)^(1/2))du$

$=int(3sec^2u)/(3secu)du$

$=intsecudu$

$=ln|secu+tanu|+c$

$=ln|sqrt(tan^2u+1)+tanu|+c$ , where, $color(blue)(tanu=x/3$

$:.I=ln|sqrt(x^2/9+1)+x/3|+c$

$=ln|sqrt(x^2+9)/3+x/3|+c$

$=ln|(sqrt(x^2+9)+x)/3|+c$

$=ln|sqrt(x^2+9)+x|-ln3+c$

$=ln|x+sqrt(x^2+9)|+C,where, C=c-ln3$