Indefinite integrals with rati0nal and polynomial functions and Substituion

Question

I am totally confused with the substitution method of evaluating indefinite integrals, especially those with rational functions and polynomials. I have 2 cases, which if I made to understand, would break ice for my future cases.

case 1: Evaluate: $\int x\sqrt{x+2}dx $

case 2: Evaluate: $\int \frac{x^3-x^2+5x-5}{x-1} dx $

For case1: I thought it was simple substitution $u=(x+2)$ then having: $\int x.u^{1/2}dx $

For which I would then integrate $\int x\int u^{1/2}dx $

to get $\frac{x^2}{2}.\frac{2.u^{3/2}}{3}$ ...and then I just plug in back my $u$.

Help on these two. Thanks in advance.

Answer 1

when you replace x you've to replace x everywhere. If $u=x+2$ then $x=u-2$ and $dx= du$. This means that your integral become \begin{equation} \int (u-2) \sqrt u du = \int u \sqrt{u} du -2 \int \sqrt{u} du=\frac{2}{15}(3u-10) u^{\frac{3}{2}} \end{equation} Now you go back by replacing $u$ by $x+2$. For the second integral please note that \begin{equation} \frac{x^3-x^2+5x-5}{x-1}=\frac{x^3-x^2}{x-1}+5=\frac{x^2(x-1)}{x-1}+5=x^2+5 \end{equation} which is easy to integrate

Answer 2

Hint for 1: If $u = x+2$ then $du = dx$ and $x = u-2$. Hence your first integral becomes $$\int x\sqrt{x+2}dx = \int(u-2)\sqrt{u}\space du$$ Remember this is single variable calculus, so it doesn't make sense to evaluate an integral like $\int x\sqrt{u}\space dx$ where both $x$ and $u$ are variables; the integral needs to be cast completely in terms of one variable.

Hint for 2: Any time you are trying to integrate a ratio of polynomials, and the polynomial in the numerator has a higher degree than the one in the denominator, try polynomial division.

Answer 3

You can use exactly the same method. For instance use the substitution $2 x^2+3=u$. With this substitution you have \begin{equation} 2 x dx = \frac{1}{2} du \end{equation} And \begin{equation} x dx = \frac{1}{4} du \end{equation} Then your integral becomes: \begin{equation} \int 3 u^5 \frac{1}{4} du =\frac{u^6}{8} \end{equation} Now you replace u by $2 x^2+3$, and you get the result.