SATHEE: Indefinite Integration 1 Question 74

Indefinite Integration 1 Question 74

75. Evaluate $\int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x$ .

$(1993, 5$ M)

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Answer:

Correct Answer: 75. $\frac{1}{2} \log 6 - \frac{1}{10}$

Solution:

Let $I = \int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x$

$\begin{aligned} = \int_{2}^{3} \frac{2 x^{5} - 2 x^{3} + x^{4} + 1 + 2 x^{2}}{(x^{2} + 1) (x^{2} - 1) (x^{2} + 1)} d x \\ = \int_{2}^{3} \frac{2 x^{3} (x^{2} - 1) + {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2} (x^{2} - 1)} d x \\ = \int_{2}^{3} \frac{2 x^{3} (x^{2} - 1)}{{(x^{2} + 1)}^{2} (x^{2} - 1)} d x + \int_{2}^{3} \frac{{(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2} (x^{2} - 1)} d x \\ = \int_{2}^{3} \frac{2 x^{3}}{{(x^{2} + 1)}^{2}} d x + \int_{2}^{3} \frac{1}{(x^{2} - 1)} d x \\ \Rightarrow I & = I_{1} + I_{2} \end{aligned}$

Now, $I_{1} = \int_{2}^{3} \frac{2 x^{3}}{{(x^{2} + 1)}^{2}} d x$

Put $x^{2} + 1 = t \Rightarrow 2 x d x = d t$

$\begin{aligned} ∴ I_{1} & = \int_{5}^{10} \frac{(t - 1)}{t^{2}} d t = \int_{5}^{10} \frac{1}{t} d t - \int_{5}^{10} \frac{1}{t^{2}} d t \\ = [\log t]_{5}^{10} + \frac{1}{t} \\ = \log 10 - \log 5 + \frac{1}{10} - \frac{1}{5} \\ = \log 2 - \frac{1}{10} \end{aligned}$

Again, $I_{2} = \int_{2}^{3} \frac{1}{(x^{2} - 1)} d x = \int_{2}^{3} \frac{1}{(x - 1) (x + 1)} d x$

$= \frac{1}{2} \int_{2}^{3} \frac{1}{(x - 1)} d x - \frac{1}{2} \int_{2}^{3} \frac{1}{(x + 1)} d x$

$\begin{aligned} = \frac{1}{2} \log (x - 1)^{3} - \frac{1}{2} \log (x + 1) \\ = \frac{1}{2} \log \frac{2}{1} - \frac{1}{2} \log \frac{4}{3} \end{aligned}$

From Eq. (i), $I = I_{1} + I_{2}$

$\begin{aligned} = \log 2 - \frac{1}{10} + \frac{1}{2} \log 2 - \frac{1}{2} \log \frac{4}{3} \\ = \log [2 \cdot 2^{1 / 2} \frac{4^{- 1 / 2}}{3}] - \frac{1}{10} = \frac{1}{2} \log 6 - \frac{1}{10} \end{aligned}$

Indefinite Integration 1 Question 74

75. Evaluate $\int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x$ .

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Indefinite Integration 1 Question 74

75. Evaluate ∫232x5+x4−2x3+2x2+1(x2+1)(x4−1)dx.

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75. Evaluate $\int_{2}^{3} \frac{2 x^{5} + x^{4} - 2 x^{3} + 2 x^{2} + 1}{(x^{2} + 1) (x^{4} - 1)} d x$ .