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Earn +20 ptsQ: What is the integral of 1 divided by 2x squared?
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How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
How do you evaluate the integral of 2udu divided by 1 plus u squared?
for the integral of (2x)dx/(1+x2 ) Take (1+x2 ) as your 'u' substitution. find du, du= 2x dx use u substitution to write new integral, integral of du/u the integral of du/u= ln abs(u) + C therefore, your original problem becomes an answer with ln ln abs (1+x2) + C *abs refers to absolute value of the parentheses
-2x squared plus x plus 1 divided by x-1?
(-2x2 + x + 1) / (x - 1) = (-2x - 1)(x - 1)/(x - 1) = -2x - 1
X squared minus 2x minus 3 divided by x minus 3?
x + 1
How do you factor 4x squared minus 2x?
2x(2x - 1)
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