answers.com lacks the facilities for writing this answer well. I suggest that you go to wolframalpha.com and enter the following query:
integrate sin(x^2+1)
The integral of 1 + x2 is x + 1/3 x3 + C.
Int sqrt(1+x2)/x = sqrt(1+x2) + LN [(sqrt(1+x2) - x -1) / (sqrt(1+x2) - x +1)]
x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2
The integral of -x2 is -1/3 x3 .
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)
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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
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
the integral of ln(sin(x)) is: -x*ln|1 - e2ix| + x*ln|sin(x)| + (i/2)*(x2 + Li2(e2ix)) + C where Li2 is the second order ploylogarithmic function.
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
The integral of 2-x = 2x - (1/2)x2 + C.