Question

${\int }_0^1\frac{(3x^4+4x^3+3x^2)}{(4x^3+3x^2+2x+1{)}^2}dx$ is equal to

• A. $\frac{1}{100}$
• B. $\frac{1}{10}$
• C. $\frac{1}{20}$
• D. 0

Answer 1

The correct option is B $\frac{1}{10}$

$\begin{array}{c}I={\int }_0^1\frac{3x^4+4x^3+3x^2}{\left(4x^3+3x^2+2x+1\right)}\\ I={\int }_0^1\frac{3x^4+4x^3+3x^2}{x^6+\left(\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3}\right)}dx\\ I={\int }_0^1\frac{{32}^{-2}+4x^{-3}+3x^{-4}}{{\left(4+\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3}\right)}^2}dx\end{array}$

Now,Put

$\begin{array}{c}\Rightarrow 4+\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3}=x\\ \Rightarrow -3x^{-2}-4x^{-3}-3x^{-4}=\frac{dt}{dx}\\ \Rightarrow \left(3x^{-2}+4x^{-3}+3x^{-4}\right)dx=-dt\end{array}$

Again,

$\begin{array}{c}I={\int }_0^1\frac{-dt}{t^2}\\ \Rightarrow {{\left[\frac{1}{t}\right]}^1}_0\\ \Rightarrow {{\left(\frac{1}{4+\frac{3}{x^2}+\frac{2}{x^3}}\right)}^1}_0\\ \left(\frac{1}{4+3+2+1}-\frac{1}{\infty }\right)\\ \Rightarrow \frac{1}{10}\end{array}$

Hence,

Option $\left(B\right)$ is correct answer