∫ x^5 dx/(x²+x+x1)(x^6+1)(x^4-x³+x-1) =

Correct Answer is : 112log⁡e∣x6−1x6+1∣1/12 _{e} ≤ft| x^6-1/x^6+1 |121logex6+1x6−1 + c

1/12 _{e} ≤ft| x^6-1/x^6+1 | + c

1/12 _{e} ≤ft| x^4+1/x^4-1 | + c

Correct Answer is : 112log⁡e∣x6−1x6+1∣1/12 _{e} ≤ft| x^6-1/x^6+1 |121logex6+1x6−1 + c

Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

\frac{x^5 dx}{(x^2+x+x1)(x^6+1)(x^4-x^3+x-1)}

Solution:

Consider the given integral

\int\;\frac{x^{5} dx}{(x^{2}+x+1)(x^{6}+1)(x^{4}-x^{3}+x-1)}

\;\;\;\; = \int\;\frac{x^{5} dx}{(x^{2}+x+1)(x^{6}+1)(x-1)(x^{3}+1)}

\;\;\;\; = \int\;\frac{x^{5} dx}{(x^{6}+1)(x^{3}-1)(x^{3}+1)}

\;\;\;\; = \int\;\frac{x^{5} dx}{(x^{6}+1)(x^{6}-1)}

Let

x^{6}=t

then

, 6 x^{5} dx = dt

This implies,

\int\;\frac{x^{5} dx}{(x^{6}+1)(x^{6}-1)} \;\; =\;\frac{1}{6} \int\;\frac{dt}{(t+1)(t-1)}

\;\; =\;\frac{1}{12} \log_{e} \left| \;\frac{t-1}{t+1} \right| + c

\;\; =\;\frac{1}{12} \log_{e} \left| \;\frac{x^{6}-1}{x^{6}+1} \right| + c

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