2

Problem:

Perform the following integration: $$ \int \dfrac{ dx } {x^2 + x + 1} $$

Answer: $$ \int \dfrac{ dx } {x^2 + x + 1} $$ Let $I$ be the integral we are trying to evaluate. \begin{align*} I &= \int \dfrac{ dx } {\left( x + \dfrac{1}{2}\right) ^2 + \dfrac{3}{4} } \end{align*} Now we use the substitution $u = x + \dfrac{1}{2}$. \begin{align*} I &= \int \dfrac{ du } {u^2 + \dfrac{3}{4} } \\ I &= \left( \dfrac{3}{4} \right) \int \dfrac{ du } {\left( \dfrac{4 }{ 3} \right) u^2 + 1 } \end{align*} Now let $v = \left( \dfrac{ \sqrt{3} }{2 }\right) u$. We have: \begin{align*} dv &= \left( \dfrac{ \sqrt{3} }{2 } \right) du \\ du &= \left( \dfrac{ 2 }{ \sqrt{3} } \right) dv \\ I &= \left( \dfrac{3(2)}{2\sqrt{3} }\right) \int \dfrac{ dv } { \left( \dfrac{4}{3} \right) \left( \dfrac{3}{4}\right) v^2 + 1 } \\ % I &= \left( \dfrac{3(2)}{2\sqrt{3} }\right) \int \dfrac{ dv }{ v^2 + 1 } \\ I &= \sqrt{3} \int \dfrac{ dv }{ v^2 + 1 } \\ I &= \sqrt{3} \arctan\left( v \right) + C \end{align*} Hence the answer is: $$ I = \sqrt{3} \arctan\left( \left( \dfrac{ \sqrt{3} }{2} \right) \left( x + \dfrac{1}{2} \right) \right) + C $$

However, an online integral calculator gets: $$ I = \dfrac{ 2 \arctan\left( \dfrac{ 2x+1 }{ \sqrt{3} } \right) } { \sqrt{3} } + C $$ Where did I go wrong?

Here is an updated answer. I believe it is right. I am hoping somebody can confirm that.

$$ \int \dfrac{ dx } {x^2 + x + 1} $$ Let $I$ be the integral we are trying to evaluate. \begin{align*} I &= \int \dfrac{ dx } {\left( x + \dfrac{1}{2}\right) ^2 + \dfrac{3}{4} } \end{align*} Now we use the substitution $u = x + \dfrac{1}{2}$. \begin{align*} I &= \int \dfrac{ du } {u^2 + \dfrac{3}{4} } \\ I &= \left( \dfrac{4}{3} \right) \int \dfrac{ du } {\left( \dfrac{4 }{ 3} \right) u^2 + 1 } \end{align*} Now let $v = \left( \dfrac{ 2 }{ \sqrt{3} }\right) u$. We have: \begin{align*} dv &= \left( \dfrac{ 2 }{ \sqrt{3} } \right) du \\ du &= \left( \dfrac{ \sqrt{3} }{ 2 } \right) dv \\ I &= \left( \dfrac{4}{3}\right) \left( \dfrac{ \sqrt{3} }{2}\right) \int \dfrac{ dv }{ v^2 + 1 } \\ % I &= \left( \dfrac{ 2\sqrt{3} }{3}\right) \arctan( v ) \\ % \end{align*} Hence the answer is: $$ I = \left( \dfrac{ 2\sqrt{3} }{3}\right) \arctan\left( \left( \dfrac{2}{ \sqrt{3} }\right) \left( x + \dfrac{1}{2} \right) \right) \\ $$

An online integral calculator gets: $$ I = \dfrac{ 2 \arctan\left( \dfrac{ 2x+1 }{ \sqrt{3} } \right) } { \sqrt{3} } + C $$ The answers are equivalent.

Bob
  • 3,954
  • 8
    when you pulled out the fraction $\frac{3}{4}$ from the denominator from the integral, since it was in the denominator, you should have taken out the reciprocal. So $\frac{4}{3} \int \frac{du}{\frac{4}{3}u^2+1}$ – Nuraly May 24 '23 at 12:02
  • 5
    Also, you should have $v=(2/\sqrt 3 )u$ – Eric May 24 '23 at 12:02

1 Answers1

0

You need to justify your substitution $u$, you can do that because:

$$\int \frac{dx}{(x+1/2)^2 + \frac{3}{4}} = \int \frac{1}{(x+1/2)^2 + \frac{3}{4}} \cdot 1 \; dx \quad \text{ and } \quad 1 = (x+ 1/2)'$$

After this,

$$I = \int \frac{du}{u^2 + \frac{3}{4}} = \int \frac{du}{\frac{3}{4}(\frac{4}{3}u^2+1)} = \frac{1}{\frac{3}{4}}\int \frac{du}{\frac{4}{3}u^2+1} = \frac{4}{3}\int \frac{du}{\frac{4}{3}u^2+1}$$

So,

$$\frac{4}{3}\int \frac{du}{\frac{4}{3}u^2+1} = \frac{4}{3}\int \frac{du}{\left(\frac{2u}{\sqrt{3}}\right)^2+1} = \frac{4}{3} \arctan{\frac{2u}{\sqrt{3}}} + c, c \in \mathbb{R}$$

Make the substitution initial $u= x+ \frac{1}{2}$,

$$ = \frac{4}{3} \arctan{\frac{2 (x + \frac{1}{2})}{\sqrt{3}}} + c = \frac{4}{3} \arctan{\frac{2 \sqrt{3} x + 1}{3}} + c, c \in \mathbb{R}$$

Dtinas10
  • 45