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If $$f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0),$$ and $$f(0)=0,$$ then the value of $$f(1)$$ is（ ）
A. $$-\frac{1}{2}$$
B. $$-\frac{1}{2}$$
C. $$\frac{1}{4}$$
D. $$\frac{1}{2}$$  Speed
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04:51 QuestionMathsClass 12

If $$f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0),$$ and $$f(0)=0,$$ then the value of $$f(1)$$ is（ ）
A. $$-\frac{1}{2}$$
B. $$-\frac{1}{2}$$
C. $$\frac{1}{4}$$
D. $$\frac{1}{2}$$

C
4.6     4.6     Solution

We have, $$f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x$$
$$=\int \frac{5\left(\frac{x^{8}}{x^{14}}\right)+7\left(\frac{x^{6}}{x^{14}}\right)}{\left(\frac{x^{2}}{x^{7}}+\frac{1}{x^{7}}+\frac{2 x^{7}}{x^{7}}\right)^{2}} d x$$
(dividing both numerator and denominator by $$x^{14}$$ )
$$=\int \frac{5 x^{-6}+7 x^{-8}}{\left(x^{-5}+x^{-7}+2\right)^{2}} d x$$
Let $$x^{-5}+x^{-7}+2=t$$
$$\Rightarrow\left(-5 x^{-6}-7 x^{-8}\right) d x=d t$$
$$\Rightarrow \quad\left(5 x^{-6}+7 x^{-8}\right) d x=-d t$$
$$\therefore f(x)=\int-\frac{d t}{t^{2}}=-\int t^{-2} d t$$
$$=-\frac{t^{-2+1}}{-2+1}+C=-\frac{t^{-1}}{-1}+C=\frac{1}{t}+C$$
$$=\frac{1}{x^{-5}+x^{-7}+2}+C=\frac{x^{7}}{2 x^{7}+x^{2}+1}+C$$
$$\because f(0)=0$$
$$\therefore 0=\frac{0}{0+0+1}+C \Rightarrow C=0$$
$$\therefore f(x)=\frac{x^{7}}{2 x^{7}+x^{2}+1}$$
$$\Rightarrow f(1)=\frac{1}{2(1)^{7}+1^{2}+1}=\frac{1}{4}$$          