Practice Question On Indefinite integration For NDA , Airforce and IITJEE | 31-45 | CBSE | SAT | COOP

Indefinite integration

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Question no. 31

What is ∫ x/(1-xcotx) dx is equal to ?

looks_one log(cosx- xsinx) + c

looks_two log(xsinx-cosx) + c

looks_3 log(sinx - xcosx) + c

looks_4 none of these

Option looks_3 log(sinx - xcosx) + c

Solution :

log(sinx - xcosx) + c

Question no. 32

What is ∫ (x-2)/(x²-4x+3) dx equal to ?

looks_one log√(x²-4x+3) + c

looks_two xlog(x-3) - 2log(x-2) +c

looks_3 log(x-3)(x-1) +c

looks_4 none of these

option looks_one log√(x²-4x+3) + c

Solution :
∫ (x-2)/(x²-4x+3) dx
Let x²-4x+3 = t
Differentiate w.r.t x
(2x-4)dx = dt
Put the value of dx in following integration
∫ 1/2t dt
(1/2)log(t) + c
log(t)^1/2 + c
log(x²-4x+3)^1/2 + c

Question no. 33

What is ∫ 3x²/(x⁶ +1) dx equal to ?

looks_one log(1+x⁶) + c

looks_two tan^-1(x³) + c

looks_3 3tan^-1(x³) + c

looks_4 3tan^-1(x³/3) + c

Option looks_two tan^-1(x³) + c

Solution :
∫ 3x²/(x⁶ +1) dx
let x³ = t
Differentiate w.r.t x
3x² dx = dt
Put the value of dx in following integration
∫ 1/(t² + 1) dt
tan^-1(t) + c
tan^-1(x³) + c

Question no. 34

What is ∫ 1/(e^x-1) dx equal to ?

looks_one ln(1 - e^-x) + k

looks_two -ln(1 - e^-x) + k

looks_3 ln(e^x-1) + k

looks_4 none of these + k

Option looks_one ln(1 - e^-x) + k

Solution :
∫ 1/(e^x-1) dx
let e^x-1 = t
Differentiate w.r.t x
e^x dx = dt
∫ 1/t(e^x) dt
∫ 1/t(t+1) dt = ∫ 1/t - 1/(t+1) dt
ln(t) -ln(t+1) + c
ln(e^x-1) - ln(e^x) + c
ln((e^x-1)/e^x) + c
ln(1 - e^-x) + c

Question no. 35

What is ∫ cos√x /√x dx equal to ?

looks_one 0.5cos√x + c

looks_two 2cos√x + c

looks_3 sin√x + c

looks_4 2sin√x + c

Option looks_4 2sin√x + c

Solution :
∫ cos√x /√x dx
√x = t
Differentiate w.r.t x
dx = 2√xdt
dx = 2t dt
∫ (2t/t)cost dt = ∫ 2cost dt
2sint + c
2sin√x + c

Question no. 36

What is ∫ cos2x/(sinx+cosx)² dx equal to ?

looks_one log√(sinx+cosx) + c

looks_two log(sinx - cosx) + c

looks_3 log(sinx + cosx) + c

looks_4 -1/(sinx+cosx) + c

option looks_3 log(sinx + cosx) + c

Solution :

let sinx + cosx = t
differentiate w.r.t x
(cosx -sinx )dx =dt
∫ 1/t dt
log(t) + c
log(sinx + cosx) + c

Question no. 37

What is ∫ x cosx dx equal to ?

looks_one xsinx +cosx + c

looks_two xsinx -cosx +c

looks_3 xcosx +sinx + c

looks_4 xcosx - sinx + c

Option looks_one xsinx +cosx + c

Solution :
∫ x cosx dx
Integrate By parts
x sinx -∫sinx dx
xsinx +cosx + c

Question no. 38

What is ∫ xcos²x dx equal to ?

looks_one (1/4)x² -(1/4)xsin2x -(1/8)cos2x + c

looks_two (1/4)x² +(1/4)xsin2x +(1/8)cos2x+ c

looks_3 (1/4)x² -(1/4)xsin2x +(1/8)cos2x+ c

looks_4 (1/4)x² +(1/4)xsin2x -(1/8)cos2x + c

option looks_two (1/4)x² +(1/4)xsin2x +(1/8)cos2x+ c

Solution :

Question no. 39

∫ sin(logx) + cos(logx) dx =

looks_one xcos(logx) + c

looks_two sin(logx) + c

looks_3 cos(logx) + c

looks_4 xsin(logx) + c

Option looks_4 xsin(logx) + c

Solution :
∫ sin(logx) + cos(logx) dx
log x = t ⇒ x = e^t
Differentiate w.r.t s
dx = e^t dt
∫ sin(logx) + cos(logx) dx
e^tsint - ∫e^tcost dt + ∫e^tcost dt
e^tsint + c
xsin(logx) + c

Question no. 40

What is ∫ logx/x³ dx equal to ?

looks_one (1/4x²)(2logx - 1) +c

looks_two -(1/4x²)(2logx + 1) +c

looks_3 (1/4x²)(2logx + 1) +c

looks_4 none of these

option looks_two -(1/4x²)(2logx + 1) +c

Solution :

Question no. 41

What is ∫ xⁿlogx dx equal to ?

looks_one (1/n+1)xⁿ⁺¹{logx + 1/(n+1) } + c

looks_two (1/n+1)xⁿ⁺¹{logx + 2/(n+1) } + c

looks_3 (1/n+1)xⁿ⁺¹{2logx - 1/(n+1) } + c

looks_4 (1/n+1)xⁿ⁺¹{logx - 1/(n+1) } + c

option looks_4 (1/n+1)xⁿ⁺¹{logx - 1/(n+1) } + c

Solution :

Question no. 42

What is ∫ e^2x+logx dx equal to ?

looks_one (1/4)(2x-1)e^2x + c

looks_two (1/4)(2x+1)e^2x + c

looks_3 (1/2)(2x+1)e^2x + c

looks_4 (1/2)(2x-1)e^2x + c

option looks_one (1/4)(2x-1)e^2x + c

Solution :
∫ e^2x+logx dx = ∫e^2x .e^logx dx
∫e^2x.x dx
Integrate By parts
xe^2x/2 - e^2x/4 + c

Question no. 43

What is ∫ logx log(x+2) dx equal to ?

looks_one x(logx)²+ c

looks_two x(1+logx)² + c

looks_3 x[1+(logx)²] + c

looks_4 none of these

option

very son 13

Question no. 44

What is ∫[ 1/logx - 1/(logx)²] dx equal to ?

looks_one 1/logx + c

looks_two x/logx + c

looks_3 x/(logx)²+ c

looks_4 none of these

option looks_two x/logx + c

Solution :

Question no. 45

What is ∫ logx/(1+logx)² equal to ?

looks_one 1/(1+ logx) + c

looks_two x/(1+logx)² + c

looks_3 x/(1+logx) +c

looks_4 1/(1+logx)² + c

Option looks_3 x/(1+logx) +c

Solution :
∫ (logx)/(1+logx)² dx
Let 1 + logx = t ⇒ x = e^t-1
(1/x)dx = dt
dx = xdt

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