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Integration of x^4/(x-1)(x^2+1)?

Kavya , 8 Years ago
Grade 12
anser 3 Answers
Harsh Patodia

Last Activity: 8 Years ago

PFA
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mycroft holmes

Last Activity: 8 Years ago

We can reduce the effort substantially......
\frac{x^4}{(x-1)(x^2+1)} = \frac{x^4-1+1}{(x-1)(x^2+1)}
= \frac{x^4-1}{(x-1)(x^2+1)}+ \frac{1}{(x-1)(x^2+1)}
=x+1+ \frac{1}{2} \left( \frac{1}{x-1} + \frac{x+1}{x^2+1}\right )
This is easy to integrate

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the attached solution to your problem.
\frac{x^4}{(x-1)(x^2+1)} = \frac{x^4-1+1}{(x-1)(x^2+1)} = \frac{x^4-1}{(x-1)(x^2+1)}+ \frac{1}{(x-1)(x^2+1)}
=x+1+ \frac{1}{2} \left( \frac{1}{x-1} + \frac{x+1}{x^2+1}\right )
Now integraing, we get,
x2/2 + x + 1/2( ln(x – 1) + 1/2ln(x2 + 1) + tan-1(x))
 
Hope it helps.
Thanks and regards,
Kushagra

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