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

4 Answers
Jun 11, 2018

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

Jun 11, 2018

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

Jun 11, 2018

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]

Jun 11, 2018

=-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