Question

integral of dx/1+4x^2​

Answer 1

Answer:

$x + 4 \frac{ {x}^{3} }{3} + c$

Step-by-step explanation:

GIVEN:-

$∫ 1 + 4 {x}^{2} \: \: dx$

SOLUTION:-

$= ∫1dx  + ∫ 4 {x}^{2}$

$= x + 4 \frac{ {x}^{3} }{3} + c$

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:\int\:\dfrac{1}{1\:+\:4x^2}\:dx\:=\:\dfrac{\arctan\:(\:2x\:)}{2}\:+\:C\:}}}$

Step-by-step-explanation:

The given integrand is

$\displaystyle{\sf\:\dfrac{1}{1\:+\:4x^2}}$

We have to integrate it with respect to x.

$\displaystyle{\sf\:\int\:\dfrac{1}{1\:+\:4x^2}\:dx}$

Let this integral be I.

$\displaystyle{\therefore\:\sf\:I\:=\:\int\:\dfrac{1}{1\:+\:4x^2}\:dx}$

$\displaystyle{\implies\sf\:I\:=\:\int\:\dfrac{1}{(\:2x\:)^2\:+\:1}\:dx}$

By substituting u = 2x and differentiating both sides w.r.t. x, we get,

$\displaystyle{\sf\:\dfrac{du}{dx}\:=\:\dfrac{d}{dx}\:(\:2x\:)}$

$\displaystyle{\implies\sf\:\dfrac{du}{dx}\:=\:2}$

$\displaystyle{\implies\:\boxed{\blue{\sf\:dx\:=\:\dfrac{1}{2}\:du\:}}}$

Now, the integral becomes

$\displaystyle{\sf\:I\:=\:\int\:\left(\:\dfrac{1}{u^2\:+\:1}\:\right)\:\dfrac{1}{2}\:du}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{1}{2}\:\int\:\dfrac{1}{u^2\:+\:1}\:du}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\int\:\dfrac{1}{u^2\:+\:1}\:du\:=\:\arctan\:u\:+\:C\:}}}$

$\displaystyle{\implies\sf\:I\:=\:\dfrac{1}{2}\:\arctan\:u}$

By re-substituting u = 2x, we get,

$\displaystyle{\implies\sf\:I\:=\:\dfrac{1}{2}\:\arctan\:(\:2x\:)}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\int\:\dfrac{1}{1\:+\:4x^2}\:dx\:=\:\dfrac{\arctan\:(\:2x\:)}{2}\:+\:C\:}}}}$