# integration help ! (1 Viewer)

#### natnat96

##### New Member
heyyy if anyone is bored and wants to help !

∫x^3/(2x+1) dx

i tried dividing the thing first but i didnt get the right answer :/

#### HeroicPandas

##### Heroic!
$\bg_white \int \frac{x^3}{2x+1}dx \\ Let u = 2x +1, du = xdx \\ \\ = \int \frac{x^2}{u}du \\=\int \frac{(\frac{u-1}{2})^2}{u}du \\ = \frac{1}{4}\int \frac{u^2 - 2u+1 }{u}du \\ = \frac{1}{4} \int (u - 2 + \frac{1}{u})du \\ = \frac{1}{4}\left ( \frac{u^2}{2} - 2u + ln(u)\right )+C \\ = \frac{1}{4}\left ( \frac{(2x+1)^2}{2} - 2(2x+1) + ln(2x+1)\right )+C$

#### Sy123

##### This too shall pass
heyyy if anyone is bored and wants to help !

∫x^3/(2x+1) dx

i tried dividing the thing first but i didnt get the right answer :/
$\bg_white \int \frac{x^3}{2x+1} \ dx$

By doing long division, we can see that:

$\bg_white \int \frac{x^3}{2x+1} \ dx = \int \frac{(2x+1)(\frac{x^2}{2} - \frac{x}{4} + \frac{1}{8}) - \frac{1}{8}}{2x+1} \ dx$

$\bg_white = \int \frac{x^2}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8} \cdot \frac{1}{2x+1} \ dx$

$\bg_white = \frac{x^3}{6} - \frac{x^2}{8} + \frac{x}{8} - \frac{1}{16} \ln(2x+1) + c$

#### natnat96

##### New Member
ohhh jokes my long division was wrong and this is a definite integral in my textbook. i had a 2 instead of a 4 nek minut...hahahah
thanks guys !!

#### asianese

##### Σ
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:

$\bg_white \frac{x^3}{2x+1}=Ax^2+Bx+C + \frac{R}{2x+1} where R is the remainder, and it is still divided by the divisor, and A, B, C are constants. \\ This is just the division transformation: \\P(x)=D(x)Q(x)+R(x) \Longrightarrow \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}.\\ Multiplying both sides by 2x+1 will yield: \\ x^3=(Ax^2+Bx+C)(2x+1)\\=2Ax^3+Ax^2+2Bx^2+Bx+2Cx+C+R\\=2Ax^3+x^2(A+2B)+x(B+2C)+C+R.\\ Now we equate the coefficients of powers of x on both sides, starting with x^3, x^2... and so on:\\ 2A=1 \Rightarrow A=\frac{1}{2}\\A+2B=0 \Rightarrow B=-\frac{1}{4}\\ B+2C=0 \Rightarrow C=\frac{1}{8}\\ In the constant, C+R=0 \Rightarrow R=-\frac{1}{8}.\\ Now we can write the quotient as: \\ \frac{x^3}{2x+1} = \frac{x^2}{2}-\frac{x}{4}+\frac{1}{8}-\frac{1}{8(2x+1)}.$

It is a good alternative to lengthy and sometimes troublesome long divisions.

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#### johnnysk

##### New Member
$\bg_white \int \frac{x^3}{2x+1}dx \\ Let u = 2x +1, du = xdx \\ \\ = \int \frac{x^2}{u}du \\=\int \frac{(\frac{u-1}{2})^2}{u}du \\ = \frac{1}{4}\int \frac{u^2 - 2u+1 }{u}du \\ = \frac{1}{4} \int (u - 2 + \frac{1}{u})du \\ = \frac{1}{4}\left ( \frac{u^2}{2} - 2u + ln(u)\right )+C \\ = \frac{1}{4}\left ( \frac{(2x+1)^2}{2} - 2(2x+1) + ln(2x+1)\right )+C$
Doesnt differentiating u=2x +1
Be du=2 dx not du = x dx

#### asianese

##### Σ
^ yes, his working is incorrect.

#### HeroicPandas

##### Heroic!
^ yes, his working is incorrect.
yes im always wrong on the internet

#### Capt Rifle

##### Member
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:

$\bg_white \frac{x^3}{2x+1}=Ax^2+Bx+C + \frac{R}{2x+1} where R is the remainder, and it is still divided by the divisor, and A, B, C are constants. \\ This is just the division transformation: \\P(x)=D(x)Q(x)+R(x) \Longrightarrow \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}.\\ Multiplying both sides by 2x+1 will yield: \\ x^3=(Ax^2+Bx+C)(2x+1)\\=2Ax^3+Ax^2+2Bx^2+Bx+2Cx+C+R\\=2Ax^3+x^2(A+2B)+x(B+2C)+C+R.\\ Now we equate the coefficients of powers of x on both sides, starting with x^3, x^2... and so on:\\ 2A=1 \Rightarrow A=\frac{1}{2}\\A+2B=0 \Rightarrow B=-\frac{1}{4}\\ B+2C=0 \Rightarrow C=\frac{1}{8}\\ In the constant, C+R=0 \Rightarrow R=-\frac{1}{8}.\\ Now we can write the quotient as: \\ \frac{x^3}{2x+1} = \frac{x^2}{2}-\frac{x}{4}+\frac{1}{8}-\frac{1}{8(2x+1)}.$

It is a good alternative to lengthy and sometimes troublesome long divisions.
That's a smart alternative actually....

#### nightweaver066

##### Well-Known Member
That's a smart alternative actually....
It seems so lengthy though...

$\bg_white \int \frac{x^3}{2x+1}dx$

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

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

$\bg_white =\int \frac{1}{2}x^2 - \frac{1}{4}x + \frac{1}{8}\times \frac{2x+1 - 1}{2x+1}dx$

$\bg_white =\int \frac{1}{2}x^2 - \frac{1}{4}x + \frac{1}{8} - \frac{1}{8} \times \frac{1}{2x+1}dx$

$\bg_white = \frac{1}{6}x^3 - \frac{1}{8}x^2 + \frac{1}{8}x - \frac{1}{16}ln(2x+1) + c$

Essentially long division without the long division.

#### asianese

##### Σ
It seems so lengthy though...

$\bg_white \int \frac{x^3}{2x+1}dx$

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

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

$\bg_white =\int \frac{1}{2}x^2 - \frac{1}{4}x + \frac{1}{8}\times \frac{2x+1 - 1}{2x+1}dx$

$\bg_white =\int \frac{1}{2}x^2 - \frac{1}{4}x + \frac{1}{8} - \frac{1}{8} \times \frac{1}{2x+1}dx$

$\bg_white = \frac{1}{6}x^3 - \frac{1}{8}x^2 + \frac{1}{8}x - \frac{1}{16}ln(2x+1) + c$

Essentially long division without the long division.
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).