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# A little problem

$\large \int \frac 1{x^\frac 13 + x^\frac 14} + \frac {\ln (1+x^\frac 16 )}{x^\frac 13 (1+x^\frac 32)} dx$

Note by Brilliant Member
1 year ago

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$$\text{I don't think It is possible to calculate the integration} \\ \text{The most that we can do is to break it into two parts} \\ I=I_1+I_2 \\ \text{Where } I_1 = \displaystyle\int \dfrac{1}{x^{\frac13}+x^{\frac14}}\cdot dx \quad \quad \text{Substitute } x=t^{12} \implies dx=12t^{11}dt \\ I_1 = 12\displaystyle\int \dfrac{t^8}{t+1}\cdot dt \quad \quad \text{Which can be further solved by long division to give us our answer} \\ I_1 = 12\displaystyle\int \left( t^7-t^6+t^5-t^4+t^3-t^2+t-1+\dfrac{1}{t+1} \right) \cdot dt \\ \text{The main problem is with the second part } I_2 \\ \text{I considered Integration by parts but it is still very complex}$$ · 12 months ago

bro can you tell me how did you write t^8/t+1 to t^7-t^6+t^5................ . i do by other methods( by doing +1-1 again and again) which are very long, can you tell me this ""long division method"". thank you · 12 months ago

$$\text{Doing }+1-1 \text{ is actually much better here } \\ \dfrac{t^8}{t+1} = \dfrac{t^8-1}{t+1}+\dfrac{1}{t+1} \quad \quad ,t^8-1=(t+1)(t-1)(t^2+1)(t^4+1) \\ \dfrac{t^8}{t+1} = (t-1)(t^6+t^4+t^2+1)+\dfrac{1}{t+1}$$ · 12 months ago

still what is this ""long division method"". · 12 months ago