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Just use the power rule for each part, and add or substract. The answer is y + 7y2/2 - y3/3 + C
Just use the power rule for each part, and add or substract. The answer is y + 7y2/2 - y3/3 + C
Just use the power rule for each part, and add or substract. The answer is y + 7y2/2 - y3/3 + C
Just use the power rule for each part, and add or substract. The answer is y + 7y2/2 - y3/3 + C
Wiki User
∙ 14y ago
Wiki User
∙ 14y agoJust use the power rule for each part, and add or substract. The answer is y + 7y2/2 - y3/3 + C
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Integral of 1 plus x2?
The integral of 1 + x2 is x + 1/3 x3 + C.
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
What is the integral of 1 over x squared plus 1?
arctan(x)
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
How do you integrate the integral of quantity of x plus 2 divided by x squared plus x plus 1?
3
Integrate1 plus x divided by x2 plus 1?
(1+x)/(x^2+1) Let x^2+1 =u 2x dx = du x dx = du/2 (1+x) / (x^2+1) = 1/(x^2+1) + x / (x^2+1) Integral of x dx / (x^2+1) = (1/2) integral du / u = 1/2 ln|u| --(1) Integral of 1 / (x^2+1) = arctan(x) --(2) Adding (1) and (2) Integral (1+x)/(x^2+1) = (1/2) ln(x^2+1) + arctan(x) + C
What is the integral of 1 divided by 2x squared?
0.5
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
What is the integral of 2x plus seven divided by x square plus 2x plus 5?
Hopefully I did this one correctly, if anyone sees an error please correct it. This is the problem:∫(2x+7)/(x2+2x+5)I rewrote the integral as:2∫x/(x2+2x+5) + 7∫1/(x2+2x+5)Both of these parts of the integral is in a form that should be listed in most integral tables in a calculus text book or on-line. From these tables the integral is the following:2*[(1/2)ln|x2+2x+5| - (1/2)tan-1((2x+2)/4)] + 7*[(1/2)tan-1((2x+2)/4)]Combining like terms gives the following:ln|x2+2x+5| + (5/2)*tan-1((2x+2)/4)
Integral of 1 divided by x cubed?
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
What is the integral of 2 times x divided by the quantity x plus 1 to the third power?
3
How do I round up to the next largest whole number in C plus plus?
It's either floor(x)+1 or ceil(x) depending on what you want to get for integral x numbers:x or x+1
