finding an indefinite integral of a fraction

Question

(a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.

(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$

For (a) I found that $B=2$ and $A=2$

And I am not quite sure how to integrate. I tried to split them into two $\displaystyle\int\frac{2}{x+2 }dx$ and $\displaystyle\int\frac{1-2x}{x^2+1}dx$ but I don't know how to do after.

Answer 1

Use $$\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C.$$ Note that $$\int\frac{2}{x+2}dx=2\int\frac{(x+2)'}{x+2}dx$$ $$\int\frac{1-2x}{x^2+1}dx=-\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx=-\int\frac{(x^2+1)'}{x^2+1}dx+\int\frac{1}{x^2+1}dx.$$ Here, set $x=\tan\theta$ for $$\int\frac{1}{x^2+1}dx.$$

Answer 2

Hint:

Let $u=x+2$ for \begin{equation} \int\frac{2}{x+2}dx \end{equation} Split the latter integral into two parts \begin{equation} \int\frac{1-2x}{x^2+1}dx=\int\frac{1}{x^2+1}dx-\int\frac{2x}{x^2+1}dx \end{equation} then let $x=\tan\theta$ also use identity $\sec^2\theta=1+\tan^2\theta$ for \begin{equation} \int\frac{1}{x^2+1}dx \end{equation} and let $v=x^2+1$ for \begin{equation} \int\frac{2x}{x^2+1}dx \end{equation}

Answer 3

Well, $$\int\frac{2}{x+2} dx= 2\log(x+2)$$ So you have that part. But for: $$\int\frac{1-2x}{x^2 + 1} dx$$ You must further decompose this fraction into partial fractions: $$\frac{1 - 2x}{x^2 + 1} = \frac{1}{x^2 + 1} - \frac{2x}{x^2 + 1}$$ So this integral becomes:$$\int\frac{1 - 2x}{x^2 + 1} dx= \int\frac{1}{x^2 + 1}dx - \int\frac{2x}{x^2 + 1}dx$$ $$= \tan^{-1}x - \log(x^2 + 1)$$ And overall: $$\int\frac{4-3x}{(x+2)(x^2 + 1)}= 2\log(x+2) + \tan^{-1}x - \log(x^2 + 1) + c$$ Note that the constant is ommited till the end, just to simplify the answer.