I need help please?!!!

Mathematics

2 answers:

kirill [66]2 years ago

8 0

It would unlikely be blue, but you never know what you’ll get

EastWind [94]2 years ago

4 0

Answer:unlikely

Step-by-step explanation:

1/6 is the chance for getting blue

1/6=0.16....

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Perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integra

worty [1.4K]

Answer:

$x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C$

Step-by-step explanation:

Given the integrand $\int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx$ , before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

Hence the partial form of the function $\dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}$

Integrating its partial sum

$\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx = \int\limits (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2} \int\limits{\frac{1}{x} } \, dx + \frac{2}{13} \int\limits{\frac{1}{x-2} } \, dx$