Definite Integration Question 73
Question 73
- Evaluate $\int_{2}^{3} \frac{2 x^{5}+x^{4}-2 x^{3}+2 x^{2}+1}{\left(x^{2}+1\right)\left(x^{4}-1\right)} d x$.
$(1993,5$ M)
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Answer:
Correct Answer: 75. $\frac{1}{2} \log 6-\frac{1}{10}$
Solution:
- Let $I=\int_{2}^{3} \frac{2 x^{5}+x^{4}-2 x^{3}+2 x^{2}+1}{\left(x^{2}+1\right)\left(x^{4}-1\right)} d x$
$$ \begin{aligned} & =\int_{2}^{3} \frac{2 x^{5}-2 x^{3}+x^{4}+1+2 x^{2}}{\left(x^{2}+1\right)\left(x^{2}-1\right)\left(x^{2}+1\right)} d x \ & =\int_{2}^{3} \frac{2 x^{3}\left(x^{2}-1\right)+\left(x^{2}+1\right)^{2}}{\left(x^{2}+1\right)^{2}\left(x^{2}-1\right)} d x \ & =\int_{2}^{3} \frac{2 x^{3}\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{2}\left(x^{2}-1\right)} d x+\int_{2}^{3} \frac{\left(x^{2}+1\right)^{2}}{\left(x^{2}+1\right)^{2}\left(x^{2}-1\right)} d x \ & =\int_{2}^{3} \frac{2 x^{3}}{\left(x^{2}+1\right)^{2}} d x+\int_{2}^{3} \frac{1}{\left(x^{2}-1\right)} d x \ \Rightarrow I & =I_{1}+I_{2} \end{aligned} $$
Now, $\quad I_{1}=\int_{2}^{3} \frac{2 x^{3}}{\left(x^{2}+1\right)^{2}} d x$
Put $x^{2}+1=t \Rightarrow 2 x d x=d t$
$$ \begin{aligned} \therefore \quad I_{1} & =\int_{5}^{10} \frac{(t-1)}{t^{2}} d t=\int_{5}^{10} \frac{1}{t} d t-\int_{5}^{10} \frac{1}{t^{2}} d t \ & =[\log t]_{5}^{10}+\frac{1}{t} \ & =\log 10-\log 5+\frac{1}{10}-\frac{1}{5} \ & =\log 2-\frac{1}{10} \end{aligned} $$
Again, $\quad I_{2}=\int_{2}^{3} \frac{1}{\left(x^{2}-1\right)} d x=\int_{2}^{3} \frac{1}{(x-1)(x+1)} d x$
$$ =\frac{1}{2} \int_{2}^{3} \frac{1}{(x-1)} d x-\frac{1}{2} \int_{2}^{3} \frac{1}{(x+1)} d x $$
$$ \begin{aligned} & =\frac{1}{2} \log (x-1)^{3}-\frac{1}{2} \log (x+1) \ & =\frac{1}{2} \log \frac{2}{1}-\frac{1}{2} \log \frac{4}{3} \end{aligned} $$
From Eq. (i), $I=I_{1}+I_{2}$
$$ \begin{aligned} & =\log 2-\frac{1}{10}+\frac{1}{2} \log 2-\frac{1}{2} \log \frac{4}{3} \ & =\log \left[2 \cdot 2^{1 / 2} \frac{4^{-1 / 2}}{3}\right]-\frac{1}{10}=\frac{1}{2} \log 6-\frac{1}{10} \end{aligned} $$
