Correct Answer is : 14log⁡(x4x4+1)+C1/4 ≤ft( {x^{4}}{x^{4}+1} ) + C41log(x4+1x4)+C

1/4 ≤ft( {x^{4}+1}{x^{4}} ) + C

1/4 ≤ft( {x^{4}}{x^{4}+1} ) + C

1/4 ≤ft( {x^{4}}{x^{4}+2} ) + C

Correct Answer is : 14log⁡(x4x4+1)+C1/4 ≤ft( {x^{4}}{x^{4}+1} ) + C41log(x4+1x4)+C

Test Index

TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

Question : 44 of 160

Marks: +1, -0

\int \frac{dx}{x(x^4+1)} =

Solution:

Consider the given integral

\int\;\; \frac{dx}{x(x^{4}+1)} = \int\;\; \frac{x^{4}+1-x^{4}}{x(x^{4}+1)} dx = \int\;\; \frac{x^{4}+1}{x(x^{4}+1)} dx - \int\;\; \frac{x^{4}}{x(x^{4}+1)} dx

= \int\;\; \frac{1}{x} dx - \int\;\; \frac{x^{3}}{x^{4}+1} dx

= \log |x| - \int\;\; \frac{x^{3}}{x^{4}+1} dx + C

Let,

x^{4}+1 = t

After differentiation this gives

4 x^{3} dx = dt

x^{3} dx = \;\; \frac{1}{4} dt

Substitute these values, we get,

\int\;\; \frac{dx}{x(x^{4}+1)} = \log |x| - \;\; \frac{1}{4} \int\;\; \frac{dt}{t} + C

= \log |x| - \;\; \frac{1}{4} \log |x^{4}+1| + C

= \;\; \frac{1}{4} \left[ \log |x^{4}| - \log |x^{4}+1| \right] + C

= \;\; \frac{1}{4} \log \left| \frac{x^{4}}{x^{4}+1} \right| + C

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