Answer to Question #131180 in Calculus for fumi

Question #131180
Integral Calculus
1. ∫ x^4-2x^3+3x^2-x+3/x^3-2x^2+3x dx
1
Expert's answer
2020-08-31T17:56:04-0400

ANSWER

"\\dfrac {1}{5}x^5 - \\dfrac {1}{2}x^4 + \\dfrac {1}{3}x^3 + x^2 - \\dfrac {3}{2x^2} + K"


SOLUTION

"\\int (x^4 - 2x^3 + 3x^2 - x + \\dfrac {3}{x^3} - 2x^2 + 3x) dx"


Simplifying the expression by collecting like terms:


"= \\int [x^4 - 2x^3 + (3x^2 - 2x^2) + (3x - x) + 3x^{-3}]dx"

"= \\int (x^4 - 2x^3 +x^2 + 2x + 3x^{-3}) dx"


Solving the integral gives:


"= \\dfrac {x^5}{5} - \\dfrac {2x^4}{4} + \\dfrac {x^3}{3} + \\dfrac {2x^2}{2} + \\dfrac {3x^{-2}}{-2} + K"



"= \\dfrac {1}{5}x^5 - \\dfrac {1}{2}x^4 + \\dfrac {1}{3}x^3 + x^2 - \\dfrac {3}{2x^2} + K"


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