Integrate $\int \frac{2x+7}{(x+1)(x^2+1)}dx$.

Question

I came across this question in a book - and I've been trying to solve it. I am only familiar with the integration using standard formulae, substitution and by parts. I'm not familiar with the other methods yet - so please show me a way to solve it using the above methods.

My work:

I couldn't find a way to substitute a function or simplify the equation in any way.

Answer 1

Using partial fractions, you can decompose the integrand into

$$ \frac{2x+7}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} $$

which gives $$ \begin{align} 2x+7 &= A(x^2+1) + (Bx+C)(x+1) \\ &= (A+B)x^2 + (B+C)x + (A+C) \end{align} $$

Comparing coefficients gives $$ A+B = 0 \\ B+C = 2 \\ A+C = 7 $$

solving the above system we obtain $A = \frac52, \ B = -\frac52, C = \frac92$

Integrating $$ \begin{align} \int \frac{2x+7}{(x+1)(x^2+1)}dx &= \frac52\int \frac{1}{x+1}dx - \frac52\int \frac{x}{x^2+1} + \frac92 \int \frac{1}{x^2+1}dx \\ &= \color{blue}{\frac52\ln|x+1| - \frac54 \ln|x^2+1| + \frac92 \arctan x + C} \end{align} $$

Answer 2

Just an observation on the fastest way to calculate the coefficients in the partial fractions decomposition: the linear system deduced from $$2x+7 = A(x^2+1) + (Bx+C)(x +1)\tag{1}$$ is comparatively simple, but may be longer to solve if there are more terms. A fast way, if irreducible factors are simple (i.e. without an exponent) consists in successively giving $x$ the values of the poles of the fraction.

Here we would have:

• $x=-1$, so that $(1)$ becomes $$5= A\cdot 2+(B+C)\cdot0=2A,\enspace\text{whence }\quad A= \frac52.$$
• $x=i$: $(1)$ becomes $$2i+7= A\cdot0+(Bi+C)(i+1)=(B+C)i+C-B,$$ whence $$\begin{cases}\phantom{-}B+C=2\\-B+C=7\end{cases}\iff B=-\frac52,\enspace C= \frac92.$$

If the denominator of the fraction has a multiple irreducible factor, say $P^r$ ($r>1$), the partial fractions decomposition has $r$ terms $$\frac{A_1}P+\dots+ \frac{A_r}{P^r}\qquad (\deg A_i<\deg P), $$ and the above method only yields $A_r$. For the other coefficients, you will have to give $x$ other values and solve a (smaller) linear system. In case $r$ is ‘big’ (say $r>3$), resorting to polynomial division by increasing powers may be faster.