∫4ex - 252ex - 5dx = Ax + Blog2ex - 5 + c, then
A = 5 and B = 3
A = 5 and B = - 3
A = - 5 and B = 3
A = - 5 and B = - 3
B.
Let I = ∫4ex - 252ex - 5dx = ∫4ex2ex - 5dx - ∫252ex - 5dx = 4∫ex2ex - 5dx - 25∫e- x2ex - 5dxPut 2ex - 5 = u and 2 - 5e- x = v⇒ 2exdx = du⇒ 5e- xdx = dv⇒ exdx = du2 and e- xdx = dv5∴ I = 4∫du2u - 25∫du5v = 2logu - 5logv + c = 2log2ex - 5 - 5log2 - 5e- x + c = 2log2ex - 5 - 5log2ex - 5ex + c = 2log2ex - 5 - 5log2ex - 5 + 5logex + c = - 3log2ex - 5 + 5x + c⇒ I = 5x - 3log2ex - 5 + cBut it is given I = Ax + Blog2ex - 5 + c∴ A = 5 and B = - 3
∫- π2π2log2 - sinx2 + sinxdx is equal to
1
3
2
0
∫x2 + 2ax + tan-1xx2 + 1dx is equal to
loga . ax + tan-1x + c
x + tan-1xlogloga + c
ax + tan-1xloga + c
logax + tan-1x + c
If ∫fxlogsinxdx = loglogsinx + c, then f(x) is equal to
cot(x)
tan(x)
sec(x)
csc(x)
∫0π2secxnsecxn +cscxndx is equal to
π2
π3
π4
π6
∫01xtan-1xdx =
π4 + 12
π4 - 12
12 - π4
- π4 - 12
If ∫19 - 16x2dx = αsin-1βx + c, then α + 1β =
712
1912
912
If ∫0π2logcosxdx = π2log12, then ∫0π2logsecxdx =
π2log12
1 - π2log12
1 + π2log12
π2log2
∫1x2 + 4x2 + 9dx = Atan-1x2 + Btan-1x3 + C, then A - B =
16
130
- 130
- 16
If x - 5x - 7dx = Ax2 - 12x + 35 + logx - 6 + x2 - 12x + 35 + C, then A =
- 1
12
- 12