Find the indefinite integral of 1 / (x ^ 4 + 16)

Find the indefinite integral of 1 / (x ^ 4 + 16)

LIM (x tends to 0): what is 5x + (SiNx) ^ 2-2x ^ 3 / TaNx + 4x ^ 2

Use the law of lobita
lim(x→0）[5x+(sinx)^2-2x^3]/(tanx+4x^2 )
=lim(x→0）[5+2sinxcosx-6x^2]/(sec^2x+8x )
=5

Limx tends to 0 (SiNx ^ 3) TaNx / 1-cosx ^ 2

It is proved that the limit of arccotx divided by X is equal to 1 when x approaches 0

Wrong, it should be
　lim(x→0)(arctanx/x)
　　= lim(x→0)(t/tant) (x = tant)
　　= lim(x→0)(t/sint)*cost
　　= 1*1 = 1.

Find the function limit: Lim [(3x-1) ^ 6 * (1-2x) ^ 4] / (3x + 5) ^ 10 x tends to infinity

lim[（3x-1）^6*(1-2x)^4]/（3x+5）^10=lim[（3-1/x）^6*(1/x-2)^4]/（3+5/x）^10=
lim[（3^6*2^4]/3^10=16/81

Find the limit [(2x-1) ^ 2] / (3x ^ 3 + 2x - 1) x tends to infinity

lim[4x²-4x+1]/(3x³+2x -1)
=lim[4/x-4/x²+1/x³]/(3+2/x² -1/x³)
=0

When x tends to infinity, find the limit of (2x ^ 3 + 3x + 1) / (4x ^ 5 + 2x + 7)

lim(x->∞)(2x^3+3x+1)/(4x^5+2x+7)
=lim(x->∞)(2/x^2+3/x^4+1/x^5)/(4+2/x^4+7/x^5)
=0

Monotone interval and limit of function y = x ^ 3-3x + 1

F '(x) = 3x ^ 2-3 Let f' (x)

Vertex coordinates of y = 1 / 2 square - 3x + 1

Take it as the form of y = ax square + BX + C
Vertex coordinate formula: the abscissa of the vertex is - B / 2a, and the ordinate takes (4a) as the denominator (4ac-b Square) as the numerator, that is (3, - 7 / 2)

The vertex of the parabola y = - half x square + 3x + 7 / 2 is a. it intersects the positive half axis of X axis at B and Y axis at C
(1) Find the coordinates of ABC
(2) Finding the area of △ ABC

Y = - x ^ 2 / 2 + 3x + 7 / 2 = - 1 / 2 * (x ^ 2-6x + 9) + 9 / 2 + 7 / 2 = - 1 / 2 (x-3) ^ 2 + 8, that is, vertex a coordinate is (3,8) y = 0, get (x-3) ^ 2 = 16, X1 = 7, X2 = - 1, intersect with the positive half axis of X at B, that is, B coordinate is (7,0) let x = 0, get y = 7 / 2, that is, C coordinate is (0,7 / 2) connected OD, then there is area s (ABC) = s (OAC) + s (OAB) -