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
\[\begin{align}
  & \Rightarrow \int{\dfrac{2x+1}{4-4x+x-{{x}^{2}}}\,dx} \\
 & \Rightarrow \int{\dfrac{2x+1}{4(1-x)+x(1-x)}\,dx} \\
\end{align}\]
\[\Rightarrow \int{\dfrac{2x+1}{(1-x)(4+x)}\,dx}……….. eq(i)\]
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,
\[\begin{align}
  & \Rightarrow \int{\left[ \dfrac{A}{(1-x)}+\dfrac{B}{(4+x)} \right]dx} \\
 & \Rightarrow \int{\dfrac{4A+Ax+B-Bx}{(1-x)(4+x)}dx} \\
\end{align}\]
\[\Rightarrow \int{\dfrac{4A+B+(A-B)x}{(1-x)(4+x)}}dx…………..eq(ii)\]
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
\[\begin{align}
  & \text{ 4}A+B+\,A-B=2+1 \\
 & \Rightarrow 5A=3 \\
\end{align}\]
\[\Rightarrow A=\dfrac{3}{5}\]
Putting the value of A in eq(iv), we get
\[\begin{align}
  & B=A-2 \\
 & \Rightarrow B=\dfrac{3}{5}-2 \\
 & \Rightarrow B=\dfrac{3-10}{5} \\
\end{align}\]
\[\Rightarrow B=-\dfrac{7}{5}\]
Putting the values of A and B in
\[\begin{align}
  & \int{\left[ \dfrac{A}{(1-x)}+\dfrac{B}{(4+x)} \right]dx} \\
 & \Rightarrow \int{\left[ \dfrac{\dfrac{3}{5}}{(1-x)}+\dfrac{\dfrac{-7}{5}}{(4+x)} \right]dx} \\
 & \Rightarrow \int{\dfrac{3}{5(1-x)}}dx-\,\int{\dfrac{7}{5(4+x)}}dx \\
 & \Rightarrow \dfrac{3}{5}\int{\dfrac{1}{(1-x)}}dx-\dfrac{7}{5}\int{\dfrac{1}{(4+x)}}dx \\
\end{align}\]
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.