Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$

Question

Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$

My attempt: $$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$

By applying polynomial division, it follows that $$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$

Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.

Substituting dx in the integral:

$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$

I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.

Thanks

Answer 1

Your last integral, $\int \frac{-x}{u(2x-2)} du$, doesn't make me happy.

You have both $x$ and $u$ in the integral. The point of the substitution is to get rid of the $x$'s and leave the integral in terms of $u$.

I'll take it from the last integral in terms of $x$, everything looks fine before and it goes astray after that.

$\int \frac{-x}{x^2-2x+1} dx = \int \frac{-x}{(x-1)^2} dx$. Let $u=x-1$ therefore $du=dx$ and $-x=-1-u$, then $\int \frac{-x}{(x-1)^2} dx=\int \frac{-1-u}{u^2} du=\int \frac{-1}{u^2} du -\int \frac{1}{u} du= \frac{1}{u}-\ln u=\frac{1}{x-1} - \ln {(x-1)}$.

Answer 2

I would let $u=x-1$ which makes the integrand $(u+1)^2(u-1)/u^2.$ After you multiply out the top you have a sum of powers of $u$.

Answer 3

HINT: write your Integrand in the form $$x- \left( x-1 \right) ^{-1}- \left( x-1 \right) ^{-2}$$

Answer 4

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

Answer 5

With a shift, $t=x-1$,

$$\int \frac{(t+1)^2(t-1)}{t^2} dt=\int\left(t+1-\frac1t-\frac1{t^2}\right)dt\\ =\frac{(x-1)^2}2+(x-1)-\ln|x-1|+\frac1{x-1}+C.$$

Answer 6

The most efficient method is probably to substitute $t := x - 1$, as I see Yves has done in his answer, as this rewrites the integrand as a sum of power functions. In you don't see this, proceeding the standard way goes as follows:

After writing the integrand as a 'proper polynomial fraction', rewrite the denominator in factored form: $$\frac{x^3-2x^2}{x^2-2x+1} = x - \frac{x}{x^2-2x+1} = x - \frac{x}{(x - 1)^2}.$$ We know from the method of partial fractions that we can put the second term in the form $$\frac{x}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} .$$ We can cross-multiply and compare like coefficients to solve for $A$ and $B$, but we can do this more quickly in this case by writing $$\frac{x}{(x - 1)^2} = \frac{(x - 1) + 1}{(x - 1)^2} = \frac{1}{x - 1} + \frac{1}{(x - 1)^2} .$$ The integrals of the two terms on the right-hand side are elementary.

Answer 7

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

$\int {x^2(x-1-1)\over(x-1)^2}dx$

$\int {x^2(x-1)-x^2\over(x-1)^2}dx$

$\int ({x^2\over(x-1)}-{x^2\over(x-1)^2})dx$

$\int {x^2\over(x-1)}dx-\int{x^2\over(x-1)^2}dx$

$t=x-1 => dt=dx$

$\int {(t+1)^2\over(t)}dt-\int{(t+1)^2\over(t)^2}dt$

$\int {(t^2+1+2t)\over(t)}dt-\int(1+{1\over t})^2dt$

$\int (t+{1\over t}+2)dt-\int(1+{1\over t})^2dt$

$\int t dt+\int{1\over t}dt+\int2dt-\int(1+{1\over t})^2dt$

${t^2\over2}+\ln t+2t-t+{1\over t}-2\ln t+C$

${(x-1)^2\over2}+(x-1)+{1\over x-1}-\ln |x-1|+C$