How to find integral : $\int\;\frac {x^2+3}{x^6(1+x^2)}dx$

Question

Question: How to find integral : $$\int\;\frac {x^2+3}{x^6(1+x^2)}dx$$ My attempt: \begin{align} \int\;\frac {x^2+3}{x^6(1+x^2)}dx＝\int\;\frac {x^2＋1＋2}{x^6(1＋x^2)}dx & ＝\int\;\frac {x^2＋1}{x^6(1＋x^2)}＋\frac {2}{x^6(1＋x^2)}dx \\ &＝\int\;\frac {1}{x^6}dx＋2\int\;\frac {1}{x^6(1＋x^2)}dx \\ &＝-\frac {1}{x^5}＋2\int\;\frac {1}{x^6(1＋x^2)}dx \end{align}

I need help calculating this integral

$$\int\;\frac {1}{x^6(1＋x^2)}dx$$

I appreciate your interest

Let $y=\frac1x$ $$\int;\frac {1}{x^6(1＋x^2)}dx=-\int \frac{(y^2+1-1)^3}{y^2+1}dy$$ — Quanto, Mar 28 '23 at 13:33
$\frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}$, now do this with functions :-) — Dominique, Mar 28 '23 at 13:41
@MathStackexchangeIsNotSoBad: you mean "enlightening", not "deceptive" :-) — Dominique, Mar 28 '23 at 13:49

score 5 · Accepted Answer · answered Mar 28 '23 at 13:35

Let $\displaystyle I =\int\frac{1}{x^6(1+x^2)}dx$ , Put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$

So $\displaystyle I =-\int\frac{t^6}{1+t^2}dt=-\int \frac{((t^2)^3+1)-1}{1+t^2}dt$

$\displaystyle I =-\int\frac{(t^2)^3+1^3}{1+t^2}dt+\int\frac{1}{1+t^2}dt$

$\displaystyle I =-\int\frac{(t^2+1)(t^4-t^2+1)}{1+t^2}dt+\int\frac{1}{1+t^2}dt$

$\displaystyle I =-\int (t^4-t^2+1)dt+\int\frac{1}{1+t^2}dt$

$\displaystyle I =-\bigg[\frac{t^5}{5}-\frac{t^3}{3}+t\bigg]+\tan^{-1}(t)+C$

score 3 · Answer 2 · answered Mar 28 '23 at 13:36

3

$$\int\;\frac {x^2+3}{x^6(1+x^2)}dx $$ $$=\int\left(-\frac{2}{x^2+1}+\frac{2}{x^2}-\frac{2}{x^4}+\frac{3}{x^6}\right)dx$$

I hope you can continue after this.

answered Mar 28 '23 at 13:36

MathStackexchangeIsNotSoBad

4,716

Ok you are lit! How did you split that integral. Could you teach me sis/bro? – Elizabeth Huffman Jun 09 '23 at 04:42
Like is there some shortcut or something? – Elizabeth Huffman Jun 09 '23 at 04:43
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@ElizabethHuffman you can refer to lai's answer...the technique is called partial fraction decomposition... it's not as simple as lai did..but yeah you can do this fast when you'll have a lot of practice – MathStackexchangeIsNotSoBad Jun 09 '23 at 09:46

score 2 · Answer 3 · edited Mar 28 '23 at 15:13

2

Hint:

\begin{aligned} & \int \frac{x\left(x+3\right)}{x^5\left(1+x^2\right)} dx\\ & \Rightarrow \int \frac{x^2+3}{x^6\left(1+x^2\right)}dx =\int \frac{1+x^2}{\left.1+x^2\right) x^6}dx+\int \frac{2}{x^6\left(1+x^2\right)} dx\\ & \Rightarrow \int \frac{1}{x^6} d x+\int \frac{x}{t^3(1+t)}d t \\ & \Rightarrow \left[ \frac{x^{-6+1}}{-6+1}\right]+\int \frac{dt}{t^{3 / 2}+t^{3 / 2}} \\ & \Rightarrow \frac{x^{-5}}{-5}+\frac{1}{2} \int t^{-3 / 2} d t \\ &= \frac{x^{-5}}{-5}+\frac{1}{2} \frac{t^{\frac{-3}{2}+1}}{-\frac{3}{2}+1} \\ & \Rightarrow \frac{x^{-5}}{5}+\frac{1}{2} \times \frac{t^{-\frac{1}{2}}}{-\frac{1}{2}} \\ & \end{aligned} After splitting the integral. I used $t=x^2$

edited Mar 28 '23 at 15:13

Max0815

3,505

answered Mar 28 '23 at 13:36

Elizabeth Huffman

808

2

I don't want to discourage you from posting answers on stackexchange by downvoting you...but here you should write your answers in Latex – MathStackexchangeIsNotSoBad Mar 28 '23 at 13:38
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@MathStackexchangeIsNotSoBad Ok thank you so much for the kindness. It's the first time I'm posting an answer. I havn't Latex notation yet. But will rewrite the entire thing in Latex for sure. +1 to your comment for being kind. – Elizabeth Huffman Mar 28 '23 at 13:41
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@ElizabethHuffman Ma'am I request you to write any question or solution using latex, as it will be searchable and also some devices do not load pictures ma'am. – Deeponjit Bose Mar 28 '23 at 14:27
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@DeeponjitBose Yes I did do as per your request:) Please do upvote if you do like my solution. Cheers! – Elizabeth Huffman Mar 28 '23 at 14:37
@MathStackexchangeIsNotSoBad Thank you once again for being kind – Elizabeth Huffman Mar 28 '23 at 14:37
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@ElizabethHuffman Yes ma'am thank you very much ma'am, and I upvoted your solution , as you have wished. Your technique works well ma'am. – Deeponjit Bose Mar 28 '23 at 14:42
@ElizabethHuffman your welcome :) – MathStackexchangeIsNotSoBad Mar 28 '23 at 16:26
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Your solution does not seem to be correct. Somehow $\arctan x$ is missing. – Ryszard Szwarc Apr 02 '23 at 17:42

score 2 · Answer 4 · answered Mar 29 '23 at 09:16

$$ \begin{aligned} \frac{x^2+3}{x^6\left(1+x^2\right)} & =\frac{3 x^2+3-2 x^2}{x^6\left(1+x^2\right)} \\ & =\frac{3}{x^6}-\frac{2}{x^4\left(1+x^2\right)} \\ & =\frac{3}{x^6}-2\left(\frac{1-x^2}{x^4}+\frac{1}{1+x^2}\right) \\ & =\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{1+x^2} \end{aligned} $$ Wish it helps to understand how to resolve the integrand!

How to find integral : $\int\;\frac {x^2+3}{x^6(1+x^2)}dx$

4 Answers4