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integration help ! (1 Viewer)
- Thread starter natnat96
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HeroicPandas
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Sy123
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asianese
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If you are uncomfortable with long division, you can try this method:
We know that a power 3 divided by power 1 will result in a power 2 polynomial with a remainder (some constant).
We can write the quotient tentatively as:
=D(x)Q(x)+R(x) \Longrightarrow \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}.\\ $Multiplying both sides by $ 2x+1$ will yield:$ \\ x^3=(Ax^2+Bx+C)(2x+1)\\=2Ax^3+Ax^2+2Bx^2+Bx+2Cx+C+R\\=2Ax^3+x^2(A+2B)+x(B+2C)+C+R.\\ $Now we equate the coefficients of powers of $x$ on both sides, starting with $x^3, x^2...$ and so on:$\\ 2A=1 \Rightarrow A=\frac{1}{2}\\A+2B=0 \Rightarrow B=-\frac{1}{4}\\ B+2C=0 \Rightarrow C=\frac{1}{8}\\ $ In the constant, $ C+R=0 \Rightarrow R=-\frac{1}{8}.\\ $ Now we can write the quotient as:$ \\ \frac{x^3}{2x+1} = \frac{x^2}{2}-\frac{x}{4}+\frac{1}{8}-\frac{1}{8(2x+1)}.)
It is a good alternative to lengthy and sometimes troublesome long divisions.
We know that a power 3 divided by power 1 will result in a power 2 polynomial with a remainder (some constant).
We can write the quotient tentatively as:
It is a good alternative to lengthy and sometimes troublesome long divisions.
Last edited:
Doesnt differentiating u=2x +1^2}{u}du \\ = \frac{1}{4}\int \frac{u^2 - 2u+1 }{u}du \\ = \frac{1}{4} \int (u - 2 + \frac{1}{u})du \\ = \frac{1}{4}\left ( \frac{u^2}{2} - 2u + ln(u)\right )+C \\ = \frac{1}{4}\left ( \frac{(2x+1)^2}{2} - 2(2x+1) + ln(2x+1)\right )+C)
Be du=2 dx not du = x dx
HeroicPandas
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yes im always wrong on the internet
Capt Rifle
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That's a smart alternative actually....If you are uncomfortable with long division, you can try this method:
We know that a power 3 divided by power 1 will result in a power 2 polynomial with a remainder (some constant).
We can write the quotient tentatively as:
It is a good alternative to lengthy and sometimes troublesome long divisions.
nightweaver066
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It seems so lengthy though...
How about this:
Essentially long division without the long division.
asianese
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That's great if you are good at spotting what coefficients to put in and balance. If you are completely lost starting from scratch (with constants to find) is perhaps a more 'direct' route, rather than fumbling around with constants (esp in exams).It seems so lengthy though...
How about this:
Essentially long division without the long division.