∫27e9x + e12x13dx is equal to
1427 + e3x13 + C
1427 + e3x23 + C
1327 + e3x43 + C
1427 + e3x43 + C
∫4dxx24 - 9x2 is equal to
4 - 9x2 + C
- 234 - 9x2 + C
- 4 - 9x2x + C
234 - 9x2x + C
∫e- xcscx1 + cotxdx is equal to
e- xcscx + C
- e- xcscx + C
- e- xcscx + cotx + C
- e- xcscx - tanx + C
∫sinx + cosxe- x + sinxdx is equal to
log1 - exsinx + C
log1 + exsinx + C
log1 - e- xsinx + C
The solution of ∫2xdtt2 - 1 = π12 is
1
2
3
4
Let f(x) = x2 - 2. If ∫36fxdx = 3fc for some c ∈ (3, 6), then the value of c is equal to
12
21
19
17
∫0π2sin2x1 + 2cos2xdx is equal to
12log2
log2
12log3
log3
∫0π2dx1 + tan3x is equal to
π
π2
π4
If f(x) = ∫1xsin2t2dt, then the value of limx→0fπ + x - fπx is equal to
14
34
D.
Given, fx = ∫1xsin2t2dtOn differentiating both sides, we get f'x = sin2x2 f'x = sin2x2ddxx - ddx1 ∵ By Leibnitz rule⇒ f'x = sin2x2∴ limx→0fπ + x - fπx = f'π = sin2π2 = 1
∫1x2x4 + 134dx is equal to
- 1 + x434x + C
- 1 + x4142x + C
- 1 + x414x + C
- 1 + x414x2 + C