∫exx2 + 1x + 12dx is equal to
exx + 1 + C
- exx + 1 + C
exx - 1x + 1 + C
exx + 1x - 1 + C
∫x + 1x1 + xexdx is equal to
log1 + xexxex + C
logxex1 + xex + C
log1 + xex + C
∫fxg'x - f'xgxfxgxloggx - logfxdx is equal to
loggxfx + C
12loggxfx2 + C
gxfxloggxfx + C
loggxfx - gxfx + C
∫0π4sinx + cosx3 + sin2xdx = ?
12log3
log2
log(3)
14log3
∫- 111 + x + x2 - 1 - x + x21 + x + x2 + 1 - x + x2dx = ?
3π2
π2
0
- 1
The area of the region described by(x, y)/x2 + y ≤ 1 and y ≤ 1 - x is
π2 - 23
π2 + 23
π2 + 43
π2 - 43
The solution of dydx + 1x = eyx2 is
2x = 1 +Cx2 ey
x = 1 +Cx2 ey
2x2 = 1 +Cx2 e - y
x2 = 1 +Cx2 e - y
A.
We have, dydx + 1x = eyx2⇒ e - ydydx + 1xe - y = 1x2Put e - y = v⇒ - e - ydydx = dvdx∴ dvdx - 1xv = - 1x2Here, above equation is a linear differential equation in v∴ IF = e∫ - 1xdx = e - logx = 1xHence, the required solution isv . 1x = ∫ - 1x2 . 1xdx + C1⇒ vx = 12x2 + C1 ⇒ vx = 1 + 2x2C12x2⇒ 2xv = 1 +Cx2 ∵ 2C1 = C⇒ 2xe - y = 1 +Cx2 ∵ v = e - y⇒ 2x = 1 +Cx2 ey
The differential equation dydx = 1ax + by + c ,where a, b, c are all non-zero real numbers, is
linear in y
linear in x
linear in both x and y
homogeneous equation
Let D be the domain of a twice differentiable function f.For all x ∈ D, f"(x) + f(x) = 0 and f(x)= ∫g(x)dx + constant.If h(x) = f(x)2 + g(x)2 and h(0) = 5, then h(2015) - h(2014) is equal to
5
3
1
The foci of the ellipse x216 + y2b2 = 1 and the hyperbola x2144 - y281 = 125 coincide. Then, the value of b2 is
7
9