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How to find the value of the following integral? Is it an integer value ?

$$\int_{1}^{2}\frac{1}{(5+3x)^{\frac{7}{4}}(4+x)^{\frac{3}{2}}}dx$$

My Approach is bit different from the question but still I am showing it.

I know how to evaluate integrals like

$$\int\frac{1}{(5+3x)(4+x)}dx$$ and $$\int\frac{1}{(5+3x)(4+3x)}dx$$.

Now let me show how can we integrate these two integrals.

For the $1^{st}$ integral, the solution is :

$$\int\frac{1}{(5+3x)(4+x)}dx=\int\frac{1}{3(\frac{5}{3}+x)(4+x)}dx=\frac{1}{3}\int\frac{A}{5+3x}dx+\frac{1}{3}\int\frac{B}{4+x}dx$$

Now this can be easily integrated using partial fraction method.

Similarly for the second integration I will write the numerator as $(5+3x)-(4+3x)$. The integration will be $$\int\frac{1}{4+3x}dx+\int\frac{1}{5+3x}dx$$

Now it is very very very very much simple and easy.

But for my question, I was thinking of multiplying the numerator and denominator by $(1+2x)$.

If I write my integration without limits, then my integration will be

$$\int\frac{1+2x}{(1+2x)(5+3x)^{\frac{7}{4}}(4+x)^{\frac{3}{2}}}dx$$

The next step of this integration is

$$\int\frac{(5+3x)-(4+x)}{(1+2x)(5+3x)^{\frac{7}{4}}(4+x)^{\frac{3}{2}}}dx$$

$$=\int\frac{1}{(1+2x)(5+3x)^{\frac{3}{4}}(4+x)^{\frac{3}{2}}}dx-\int\frac{1}{(1+2x)(5+3x)^{\frac{3}{4}}(4+x)^{\frac{1}{2}}}dx$$

Now I am facing huge problem to integrate the integration further.

My Request: Pls help.

  • See this post, how to approach such integrals, or this post. – Dietrich Burde Aug 25 '24 at 12:47
  • Your post is not matching with integration of my question. Your post will work only when $x$ is there. But in my question $3x$ and $x$ are there. –  Aug 25 '24 at 12:48
  • No need to match exactly, you still can apply the same idea. – Dietrich Burde Aug 25 '24 at 12:49
  • Ok. Let me check thoroughly @DietrichBurde –  Aug 25 '24 at 12:50
  • The post Dietrich Burde is not the exact same integral, but it's similar, so why not try to use the same ideas instead of immediately dismissing the help that is being offered to you? – jjagmath Aug 25 '24 at 12:53
  • Yes surely I am taking his help.@jjagmath –  Aug 25 '24 at 12:56
  • According to @DietrichBurde's post, if I substitute $(4+x)=u$ and $u^{\frac{1}{2}}=t$, then my integration will be $\int\frac{2}{(3t^{2}-7)^{\frac{7}{4}}(t^{2})}dt$. After this step how to proceed ? –  Aug 25 '24 at 13:08
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    The value of the integral is clearly not an integer value: When $x \geq 1$ $0 \leq \frac{1}{(5+3x)^{\frac{7}{4}}(4+x)^{\frac{3}{2}}} \leq \frac{1}{8^{\frac{7}{4}}5^{\frac{3}{2}}} \leq \frac{1}{8^1\cdot 5^1} = \frac{1}{40}$. – zetko Aug 25 '24 at 13:12
  • Hi @SubhaSankarRoy. don't be desperate, wait. Hopefully you are going to get solution here. –  Aug 25 '24 at 13:13
  • Let's write it in more elegant form $$\int_0^1 \frac{1}{(3x+2)^{7/4} , (x+3)^{3/2}} dx$$ –  Aug 25 '24 at 13:15
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    @LuckyChouhan Please correct the limits of the integral, using an other letter is always better, then the interval $[1,2]$ is mapped in the right integral by setting $t=x+1$, $x=t-1$, so for $x=1,2$ the corresponding values for $t$ are $2,3$. – dan_fulea Aug 25 '24 at 23:05
  • Which is the source of this problem? The integral can be seen as an integral over a curve of genus one... So mentioning the source and the own knowledge in domains related to the given integral do matter. The attempt is far from doing something in a direction of progress. Yes, we may use the partial fraction decomposition for $$\frac 1{(5+3x)(4+x)}$$ to get related integrals with "lower fractional powers", but then we face the same complexity. – dan_fulea Aug 25 '24 at 23:09
  • @dan_fulea thank you for pointing out my mistake! Actually what I did was $x \to x-1$ but I replaced $x$ in integrand with $x-1$ instead of $x+1$ :( –  Aug 26 '24 at 01:06

2 Answers2

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By the change of variable $$s:=\frac{3x+5}7,$$

you turn the integrand to

$$\propto(s+1)^{-7/4}s^{-3/2}.$$

You can obtain the value as an incomplete Beta integral. I doubt that there is a closed form for these exponents. WA gives an hypergeometric function.

https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function

  • $s$ can't be equal to $\frac{3(x+4)}{7}$. Because then $(s+1)=\frac{3x+19}{7}$ which is not matching with my question –  Aug 25 '24 at 13:40
  • @SubhaSankarRoy: fixed, thanks for the notice. –  Aug 25 '24 at 21:14
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Too long for a comment,

We can rewrite the given integral as \begin{align} I &= \int_0^1 \frac{1}{(3x+8)^{7/4} (x+5)^{3/2}} dx \\ &= \int_0^1 \frac{3\sqrt{3}}{(3x+8)^{7/4} (3x+15)^{3/2}} dx \\ &= \frac{3\sqrt{3}}{7} \int_0^1 \left( \underbrace{\frac{1}{(3x+8)^{7/4} (3x+15)^{1/2}}}_{I_1} - \underbrace{\frac{1}{(3x+8)^{3/4} (3x+15)^{3/2}}}_{I_2} \right) dx \end{align}

Now $I_1$ might be evaluated by as suggested in linked post. For, \begin{align} I_2 &= \frac{1}{(3x+8)^{3/4} (3x+15)^{3/2}} \\ &= \frac{1}{7} \left( \frac{1}{(3x+8)^{3/4} (3x+15)^{1/2}} - \frac{1}{(3x+8)^{-1/4} (3x+15)^{3/2}} \right) \end{align}

Now again $I_2$'s first one can be done as $I_1$ and for second one substitute $3x+8 = y^4$