(5x 2 - 8x + 1) + (2x 2 - 4x - 11)
∫1/(1-x)2dxWe can rewrite the integral as ∫(1-x)-2dx.Thus:∫(1-x)-2dxu = 1-xdu = -1-1∫-1(1-x)-2dx-1∫u-2du(u-1|0u=1-x(1-x)-1So we conclude that ∫1/(1-x)2dx = (1-x)-1 or 1/(1-x)
2(x + 3)(x + 1)
4
yes 1 + cot x^2 = csc x^2
3
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
104
1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 = 1 + 1 + 1 + 1 + 1 + 1 = 5
The indefinite integral of (1/x^2)*dx is -1/x+C.
it would be 1 fourth multiplied by quanity d squared plus 3d plus 9 (1/4)(d2 + 3d + 4)
0.5
1 squared plus 8 squared or 4 squared plus 7 squared
the integral of the square-root of (x-1)2 = x2/2 - x + C
The integral of 1 + x2 is x + 1/3 x3 + C.
x[x+1] squared Simplified is, to my knowledge, x squared plus ( x + 1) squared
ln(sinx) + 1/3ln(sin3x) + C