How to find #int(10x^3-7x)/(5x^4-7x^2+8)dx# ?

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Answer 1 · 2017-06-16T09:41:21

The answer is #=1/2ln(|5x^4-7x^2+8|)+C#

We perform this integral by substitution

Let #u=5x^4-7x^2+8#

#du=(20x^3-14x)dx#

#du=1/2(10x^3-7x)dx#

Therefore,

#int((10x^3-7x)dx)/(5x^4-7x^2+8)=1/2int(du)/u#

#=1/2lnu#

#=1/2ln(|5x^4-7x^2+8|)+C#