Math, asked by Raman786, 1 year ago

integral of x³+x+1/x²-1

Raman786: its ans is -2 under root (1-x)+cos inverse under root x+ under root x.under root (1-x) i need details

Raman786: but now my ans is x²/2+3/2 (log(x-1))+1/2(log(x+1))+c

Anonymous: Differentiate it back if you get the question then your answer is right

kvnmurty: the question is not specified properly. is it (x³+x+1 ) / (x² - 1) ?

Raman786: ok i have ans but i need ans with steps

mkt8995: X^4 /4 +x^2/2-1/x+x

kvnmurty: Mkt8995 - your answer is not right. the question is what i wrote in the comments now. please change ur answer.

Raman786: but by which way thats right ans

Answers

Answered by kvnmurty

you should use parentheses for writing proper expressions.

$\frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1}=\\ \\Integrate\ now\\ \\$

$\int\limits^{}_{} {(x+\frac{3}{2}\frac{1}{x-1}+\frac{1}{2}\frac{1}{x+1}}) \, dx \\ \\= \int\limits^{}_{} {x} \, dx + \frac{3}{2} \int\limits^{}_{} {\frac{1}{x-1}} \, dx + \frac{1}{2}\int\limits^{}_{} {\frac{1}{x+1}} \, dx \\ \\\frac{1}{2}x^2+\frac{3}{2}Ln(x-1)+\frac{1}{2}Ln(x+1)+C\\$

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