Question

How can you find the antiderivative of 1/(1+3x²) ?

Answer 1

`int dx/(1+3x^2) = 1/sqrt3 arctan(sqrt3x)+C`

Explanation:

`int dx/(1+3x^2) = int dx/(1+(sqrt3x)^2)`

`int dx/(1+3x^2) = 1/sqrt3 int (d (sqrt3x))/(1+(sqrt3x)^2)`

`int dx/(1+3x^2) = 1/sqrt3 arctan(sqrt3x)+C`

Answer 2

`1/sqrt(3)*arctan(sqrt(3)*x)+C`

Explanation:

Writing
`1/(1+(sqrt(3)*x)^2)`
we Substitute
`t=sqrt(3)x`
then we get

`dx=1/sqrt(3)dx`
and we get

`1/sqrt(3)int1/(1+t^2)dt=1/sqrt(3)*arctan(t)+C=1/sqrt(3)arctan(sqrt(3)*x)+C`

Answer 3

`int dx /(1+3 x^2)=1/sqrt3 tan^-1 sqrt3 x+ C`

Explanation:

`int 1/(1+3 x^2)dx =` Let `x = 1/sqrt 3 tan theta`

`tan theta = sqrt3 x :. theta = tan^-1 sqrt3 x`

`:. dx = 1/sqrt 3 sec ^2 theta d theta :. x ^2= 1/3 tan^2 theta`

`:. int dx /(1+3 x^2)`

`=int(1/sqrt 3 sec ^2 theta d theta)/(1+3*1/3tan^2 theta)`

`=int(1/sqrt 3 sec ^2 theta d theta)/(1+tan^2 theta)`

`=int1/sqrt 3 cancel(sec ^2 theta)/cancel(sec ^2 theta)*d theta`

`=1/sqrt3 theta+ C = 1/sqrt3 tan^-1 sqrt3 x+ C` ; [Ans]

Answer 4

`=-1/sqrt(1+3x^2)+C`

Explanation:

`int1/(1+3x^2)dx`

Integration by Trigonometric Substitution

`x=1/sqrt(3)tanu`

`dx=1/sqrt(3)tanusecu*du`

Substitute

`int1/(1+3x^2)dx=1/sqrt3int(secutanu*du)/(1+tan^2u)`

`color(green)(1+tan^2u=sec^2u`

=`1/sqrt3int(cancel(secu)tanudu)/sec^cancel(2)u`

=`1/sqrt3intsinu*du`

=`-1/sqrt3cosu+C`

reverse the substitution

`=-1/sqrt3*1/sqrt(1+3x^2)+C`