Is My Solution to This Integral Correct?

[paulmdrdo1]

again, i need some help here guys.$\displaystyle\int\frac{3x-1}{2x^2+2x+3}dx$

=$\displaystyle\int\frac{3x-1}{2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]}dx$

$\displaystyle a=\frac{\sqrt{5}}{2}$; $\displaystyle u=x+\frac{1}{2}$; $\displaystyle du=dx$; $\displaystyle x=u-\frac{1}{2}$

=$\displaystyle\frac{1}{2}\int3\frac{\left(u-\frac{1}{2}\right)}{a^2+u^2}du$;

=$\displaystyle\frac{1}{2}\int\frac{\left(3u-\frac{3}{2}\right)}{a^2+u^2}du$

=$\displaystyle\frac{1}{2}\left(\int\frac{3u}{a^2+u^2}du-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

$\displaystyle v=a^2+u^2$; $\displaystyle dv=2udu$; $\displaystyle du=\frac{1}{2u}dv$

=$\displaystyle\frac{1}{2}\left(3\int\frac{udv}{v\,2u}dv-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}\int \frac{dv}{v}-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}ln(a^2+u^2)-\frac{3}{2}\frac{2}{\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}\right)$

=$\displaystyle\frac{3}{4}ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{3}{2\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}+C$ ---> FINAL ANSWER.

would you be so generous to check if my solution is correct?

thanks!

the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is my answer equivalent to this one?

 

[Ackbach]

Re: integrals giving inverse trig.

Hmm. I have several thoughts on this.

1. You can check your answer by differentiating, and seeing if you get back to the original integrand. What does that give you?
2. My HP 50g calculator gives me
   $$\int=- \frac{ \sqrt{5}}{2} \arctan \left( \frac{ \sqrt{5}+2 \sqrt{5} x}{5} \right)+ \frac{3}{4} \ln |2x^{2}+2x+3|.$$
3. You can see for yourself what W/A gives.
4. In looking over your work, I think your first substitution might have gone awry. I wouldn't do more than one substitution at once.

 

[paulmdrdo1]

Re: integrals giving inverse trig.

what's wrong with my first substitution? please explain.

 

[alyafey22]

Re: integrals giving inverse trig.

You have $3x-1$ if we use the substitution $$x=u-\frac{1}{2}$$

we get $3\left(u-\frac{1}{2} \right)-1=3u-\frac{3}{2}-1=3u-\frac{5}{2}$

You missed the $$-1$$ in $$3x-1$$.

 

[paulmdrdo1]

Re: integrals giving inverse trig.

yes i missed that one.

my new answer is $\displaystyle\frac{3}{4}\ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this correct?

 

[alyafey22]

Re: integrals giving inverse trig.

You have the term $$\ln(a^2+u^2)$$

what do you get by letting

$$a^2=\frac{5}{4},u=x+\frac{1}{2}$$

 

[paulmdrdo1]

Re: integrals giving inverse trig.

You have the term $$\ln(a^2+u^2)$$

what do you get by letting

$$a^2=\frac{5}{4},u=x+\frac{1}{2}$$

oh hey, i forgot to square them

$\displaystyle\frac{3}{4}\ln\left(x^2+x+\frac{3}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$ will this be the correct answer?

 

[alyafey22]

Re: integrals giving inverse trig.

oh hey, i forgot to square them

$\displaystyle\frac{3}{4}\ln\left(x^2+x+\frac{3}{2}\right)-\frac{5}{2\sqrt{5}}\arctan\frac{2x+1}{\sqrt{5}}+C$ will this be the correct answer?

Yes , sure .

 

[Prove It]

again, i need some help here guys.$\displaystyle\int\frac{3x-1}{2x^2+2x+3}dx$

=$\displaystyle\int\frac{3x-1}{2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]}dx$

$\displaystyle a=\frac{\sqrt{5}}{2}$; $\displaystyle u=x+\frac{1}{2}$; $\displaystyle du=dx$; $\displaystyle x=u-\frac{1}{2}$

=$\displaystyle\frac{1}{2}\int3\frac{\left(u-\frac{1}{2}\right)}{a^2+u^2}du$;

=$\displaystyle\frac{1}{2}\int\frac{\left(3u-\frac{3}{2}\right)}{a^2+u^2}du$

=$\displaystyle\frac{1}{2}\left(\int\frac{3u}{a^2+u^2}du-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

$\displaystyle v=a^2+u^2$; $\displaystyle dv=2udu$; $\displaystyle du=\frac{1}{2u}dv$

=$\displaystyle\frac{1}{2}\left(3\int\frac{udv}{v\,2u}dv-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}\int \frac{dv}{v}-\frac{3}{2}\int\frac{du}{a^2+u^2}du\right)$

=$\displaystyle\frac{1}{2}\left(\frac{3}{2}ln(a^2+u^2)-\frac{3}{2}\frac{2}{\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}\right)$

=$\displaystyle\frac{3}{4}ln\left(x+\frac{\sqrt{5}+1}{2}\right)-\frac{3}{2\sqrt{5}}\arctan\frac{2x+1}{ \sqrt{5}}+C$ ---> FINAL ANSWER.

would you be so generous to check if my solution is correct?

thanks!

the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is my answer equivalent to this one?

A good strategy is ALWAYS to look for a simple substitution, or to see how you can manipulate your integrand to get a simple substitution...

\displaystyle \begin{align*} \int{ \frac{3x - 1}{2x^2 + 2x + 3} \, dx} &= 3\int{\frac{x - \frac{1}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x - \frac{4}{3}}{2x^2 + 2x + 3} \, dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2 - \frac{10}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \left( \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{10}{3} \int{ \frac{1}{2x^2 + 2x + 3} \, dx} \right) \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{5}{2} \int{ \frac{1}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx } - \frac{5}{4} \int{ \frac{1}{x^2 + x + \frac{3}{2} }\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx} - \frac{5}{4} \int{ \frac{1}{x^2 + x + \left( \frac{1}{2} \right) ^2 - \left( \frac{1}{2} \right) ^2 + \frac{6}{4} } \,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \, dx} - \frac{5}{4} \int{ \frac{1}{ \left( x + \frac{1}{2} \right) ^2 + \frac{5}{4} } \, dx} \end{align*}

The first of these integrals can now be solved using the substitution \displaystyle \begin{align*} u = 2x^2 + 2x + 3 \implies du = \left( 4x + 2 \right) \, dx \end{align*} and the second can be solved using the substitution \displaystyle \begin{align*} x + \frac{1}{2} = \frac{\sqrt{5}}{2} \tan{(\theta)} \implies dx = \frac{\sqrt{5}}{2} \sec^2{(\theta)}\,d\theta \end{align*}.

 

[paulmdrdo1]

hey zaid, for educational purposes could you tell me why the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this equivalent to what i get in my answer?

 

[Prove It]

Yes, they rationalised the denominator.

\displaystyle \begin{align*} \frac{5}{2\sqrt{5}} &= \frac{5}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{5\sqrt{5}}{2 \cdot 5} \\ &= \frac{\sqrt{5}}{2} \end{align*}

 

[paulmdrdo1]

A good strategy is ALWAYS to look for a simple substitution, or to see how you can manipulate your integrand to get a simple substitution...

\displaystyle \begin{align*} \int{ \frac{3x - 1}{2x^2 + 2x + 3} \, dx} &= 3\int{\frac{x - \frac{1}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x - \frac{4}{3}}{2x^2 + 2x + 3} \, dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2 - \frac{10}{3}}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \left( \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{10}{3} \int{ \frac{1}{2x^2 + 2x + 3} \, dx} \right) \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3}\,dx} - \frac{5}{2} \int{ \frac{1}{2x^2 + 2x + 3}\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx } - \frac{5}{4} \int{ \frac{1}{x^2 + x + \frac{3}{2} }\,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \,dx} - \frac{5}{4} \int{ \frac{1}{x^2 + x + \left( \frac{1}{2} \right) ^2 - \left( \frac{1}{2} \right) ^2 + \frac{6}{4} } \,dx} \\ &= \frac{3}{4} \int{ \frac{4x + 2}{2x^2 + 2x + 3} \, dx} - \frac{5}{4} \int{ \frac{1}{ \left( x + \frac{1}{2} \right) ^2 + \frac{5}{4} } \, dx} \end{align*}

The first of these integrals can now be solved using the substitution \displaystyle \begin{align*} u = 2x^2 + 2x + 3 \implies du = \left( 4x + 2 \right) \, dx \end{align*} and the second can be solved using the substitution \displaystyle \begin{align*} x + \frac{1}{2} = \frac{\sqrt{5}}{2} \tan{(\theta)} \implies dx = \frac{\sqrt{5}}{2} \sec^2{(\theta)}\,d\theta \end{align*}.

hey prove it,

can you tell me how did you arrive at 2nd line of your solution?

 

[Prove It]

hey prove it,

can you tell me how did you arrive at 2nd line of your solution?

Multiplying by 4?

 

[paulmdrdo1]

Yes, they rationalised the denominator.

\displaystyle \begin{align*} \frac{5}{2\sqrt{5}} &= \frac{5}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{5\sqrt{5}}{2 \cdot 5} \\ &= \frac{\sqrt{5}}{2} \end{align*}

oh yes, but what about the $\ln(2x^2+2x+3)$?

- - - Updated - - -

Multiplying by 4?

hmm. what made you decide that you have to multiply it by 4? wouldn't that change the integrand?

 

[Prove It]

oh yes, but what about the $\ln(2x^2+2x+3)$?

You can write it as \displaystyle \begin{align*} \ln{ \left[ 2 \left( x^2 + x + \frac{3}{2} \right) \right] } = \ln{(2)} + \ln{ \left( x^2 + x + \frac{3}{2} \right) } \end{align*}, and since \displaystyle \begin{align*} \ln{(2)} \end{align*} is a constant, it can combine with the integration constant.

hmm. what made you decide that you have to multiply it by 4? wouldn't that change the integrand?

I multiplied by 4 because I realized that if we make the simplest substitution \displaystyle \begin{align*} u = 2x^2 + 2x + 3 \end{align*} its derivative \displaystyle \begin{align*} 4x + 2 \end{align*} has to be a factor.

No, it didn't change the integrand, because I also divided by 4 outside.

 

[paulmdrdo1]

No, it didn't change the integrand, because I also divided by 4 outside.

you mean like this,

$\displaystyle 3\int\frac{\left(x+\frac{1}{3}\right)}{2x^2+2x+3} \frac{4}{4}$ ?

 

[alyafey22]

hey zaid, for educational purposes could you tell me why the answer in my book is in this form,

$\displaystyle\frac{3}{4}ln(2x^2+2x+3)-\frac{\sqrt{5}}{2}\arctan\frac{2x+1}{\sqrt{5}}+C$

is this equivalent to what i get in my answer?

This a common question and this is due to the constant of integration in that sense and since we are working on an indefinite integral let us take the following example :

$$\int \frac{1}{2x+1}\, dx $$

First approach :

$$\frac{1}{2}\int \frac{2}{2x+1}\, dx =\frac{1}{2}\ln(2x+1)+C$$

Second approach :

$$\frac{1}{2}\int \frac{1}{x+\frac{1}{2}}\, dx = \frac{1}{2}\ln\left(x+\frac{1}{2} \right) +C $$

Why the answers differ ? are my approaches correct ?