Question

Solve the expression, $\int{\dfrac{2x+1}{4-3x-{{x}^{2}}}\,dx}$

Answer 1

Hint:Express the given expression as a summation of two expressions and then calculate the integration separately. The expression can be decomposed into a sum of two expressions using factorisation. Make the deominator as the product of linear factors, assume A and B as the partial fraction in the numerator in both parts of expression respectively. Then decompose it into a sum of functions whose integrals are already known.
Complete step-by-step answer:
We can see the denominator that it can be factorized

& \Rightarrow \int{\dfrac{2x+1}{4-4x+x-{{x}^{2}}}\,dx} \\\ & \Rightarrow \int{\dfrac{2x+1}{4(1-x)+x(1-x)}\,dx} \\\ \end{aligned}$$ $$\Rightarrow \int{\dfrac{2x+1}{(1-x)(4+x)}\,dx}……….. eq(i)$$ Now, we can see that it is very difficult to integrate the given function directly. But it can be integrated after decomposing it into a sum or difference of number of functions whose integrals are already known.Make the deominator as the product of linear factors, assume A and B as the partial fraction in the numerator in both parts of expression respectively. Here, A and B are constants. Our expression will look like, $$\begin{aligned} & \Rightarrow \int{\left[ \dfrac{A}{(1-x)}+\dfrac{B}{(4+x)} \right]dx} \\\ & \Rightarrow \int{\dfrac{4A+Ax+B-Bx}{(1-x)(4+x)}dx} \\\ \end{aligned}$$ $$\Rightarrow \int{\dfrac{4A+B+(A-B)x}{(1-x)(4+x)}}dx…………..eq(ii)$$ Compare eq(i) with eq(ii),we have $$4A+B=1……………eq(iii)$$ $$A-B=2……………..eq(iv)$$ Adding eq(iii) and eq(iv),we have $$\begin{aligned} & \text{ 4}A+B+\,A-B=2+1 \\\ & \Rightarrow 5A=3 \\\ \end{aligned}$$ $$\Rightarrow A=\dfrac{3}{5}$$ Putting the value of A in eq(iv), we get $$\begin{aligned} & B=A-2 \\\ & \Rightarrow B=\dfrac{3}{5}-2 \\\ & \Rightarrow B=\dfrac{3-10}{5} \\\ \end{aligned}$$ $$\Rightarrow B=-\dfrac{7}{5}$$ Putting the values of A and B in $$\begin{aligned} & \int{\left[ \dfrac{A}{(1-x)}+\dfrac{B}{(4+x)} \right]dx} \\\ & \Rightarrow \int{\left[ \dfrac{\dfrac{3}{5}}{(1-x)}+\dfrac{\dfrac{-7}{5}}{(4+x)} \right]dx} \\\ & \Rightarrow \int{\dfrac{3}{5(1-x)}}dx-\,\int{\dfrac{7}{5(4+x)}}dx \\\ & \Rightarrow \dfrac{3}{5}\int{\dfrac{1}{(1-x)}}dx-\dfrac{7}{5}\int{\dfrac{1}{(4+x)}}dx \\\ \end{aligned}$$ Using the formula, $$\int{\dfrac{1}{x+a}dx=\ln x}$$ and $$\int{\dfrac{1}{a-x}dx=-\ln \left( a-x \right)}$$ . $$\Rightarrow -\dfrac{3}{5}\ln (1-x)-\dfrac{7}{5}\ln (4+x)+c$$ Note: In this question, one must remember the integration of $$\dfrac{1}{x+a}$$ that is, ln(x+a). By remembering this integration, the question becomes very easy to solve and also try to solve using a partial fraction method whenever the denominator part is in quadratic form.