First slide

Methods of integration

Question

If $f (x) = \int \frac{5 x^{8} + 7 x^{6}}{{(x^{2} + 1 + 2 x^{7})}^{2}} dx, (x \geq 0)$ , and $f (0) = 0,$ then the value of f(1) is

Moderate

A

$- \frac{1}{2}$
B

$- \frac{1}{4}$
C

$\frac{1}{4}$
D

$\frac{1}{2}$

Solution

We have,
$\begin{aligned} f (x) = \int \frac{5 x^{8} + 7 x^{6}}{{(x^{2} + 1 + 2 x^{7})}^{2}} dx \\ = \int \frac{5 (\frac{x^{8}}{x^{14}}) + 7 (\frac{x^{6}}{x^{14}})}{{(\frac{x^{2}}{x^{7}} + \frac{1}{x^{7}} + \frac{2 x^{7}}{x^{7}})}^{2}} dx \end{aligned}$
[dividing both numerator and denominator by x¹⁴]
$= \int \frac{5 x^{- 6} + 7 x^{- 8}}{{(x^{- 5} + x^{- 7} + 2)}^{2}} dx$
$\begin{array}{l} Let x^{- 5} + x^{- 7} + 2 = t \\ \Rightarrow (- 5 x^{- 6} - 7 x^{- 8}) dx = dt \\ \Rightarrow (5 x^{- 6} + 7 x^{- 8}) dx = - dt \\ ∴ f (x) = \int - \frac{dt}{t^{2}} = - \int t^{- 2} dt \\ = - \frac{t^{- 2 + 1}}{- 2 + 1} + C = - \frac{t^{- 1}}{- 1} + C = \frac{1}{t} + C \end{array}$
$\begin{array}{l} = \frac{1}{x^{- 5} + x^{- 7} + 2} + C = \frac{x^{7}}{2 x^{7} + x^{2} + 1} + C \\ ∵ f (0) = 0 \\ ∴ 0 = \frac{0}{0 + 0 + 1} + C \Rightarrow C = 0 \\ ∴ f (x) = \frac{x^{7}}{2 x^{7} + x^{2} + 1} \Rightarrow f (1) = \frac{1}{2 (1)^{7} + 1^{2} + 1} = \frac{1}{4} \end{array}$

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