Practice Question On Indefinite integration For NDA , Airforce and IITJEE

Indefinite integration

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Question no. 1

What is ∫ log(x+1) dx is equal to ?

looks_one xlog(x+1) -x + c

looks_two (x+1)log(x+1) -x + c

looks_3 1/(x+1) + c

looks_4 log(x+1)/(x+1) + c

Option looks_two (x+1)log(x+1) -x + c

Solution :
∫ log(x+1) dx
Integrate By parts

Question no. 2

If ∫ dx/f(x) = log {f(x)}² +c , then what is f(x) equal to ?

looks_one 2x +b

looks_two x+b

looks_3 x/2 + b

looks_4 x² + b

Option looks_3 x/2 + b

Solution :
∫ dx/f(x) = log {f(x)}² + c

Question no. 3

What is ∫ (e^-x +1 )^-1 dx equal to ?

looks_one ln(e^x +1) + c

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

looks_3 -ln(e^-x +1) + c

looks_4 -(e^x +1) + c

Option looks_one ln(e^x +1) + c

Solution :
∫ (e^-x +1 )^-1 dx

Question no. 4

What is ∫ 1/(sin²θ +2cos²θ -1) dθ equal to ?

looks_one tanθ + c

looks_two cotθ + c

looks_3 0.5tanθ + c

looks_4 0.5cotθ + c

Option looks_one tanθ + c

Solution :

Question no. 5

What is ∫ tan²xsec⁴x dx equal to ?

looks_one (1/5)sec⁵x + (1/3)sec³x + c

looks_two (1/5)tan⁵x + (1/3)tan³x + c

looks_3 (1/5)tan⁵x + (1/3)sec³x + c

looks_4 (1/5)sec⁵x + (1/3)tan³x + c

Option looks_two (1/5)tan⁵x + (1/3)tan³x + c

Solution :
∫ tan²xsec⁴x dx
tan x = t
sec²x dx = dt
∫ tan²xsec⁴x dx = ∫ t²sec⁴x dt/sec²x
∫ t²sec²x dt = ∫ t²( 1 + tan²x) dt
∫ t²( 1 + t²) dt = ∫ t² + t⁴ dt
t³/3 + t⁵/5 + C
(1/5)tan⁵x + (1/3)tan³x + c

Question no. 6

What is ∫ (a+bsinx)/cos²x dx equal to ?

looks_one asecx +btanx + c

looks_two atanx + bsecx+ c

looks_3 acotx + bcosecx + c

looks_4 acosecx + bcotx + c

Option looks_two atanx + bsecx+ c

Solution :

Question no. 7

What is ∫ (logx)/(1+logx)² dx equal to ?

looks_one 1/(1+logx)³ + c

looks_two 1/(1+logx)² +c

looks_3 x/(1+logx) +c

looks_4 x/(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

Question no. 8

What is ∫ e^lnxsinx dx equal to ?

looks_one e^lnx(sinx - cosx) + c

looks_two sinx - xcosx + c

looks_3 xsinx + cosx + c

looks_4 sinx + xcosx + c

Option looks_two sinx - xcosx + c

Solution :
∫ e^lnxsinx dx = ∫ x sinx dx
-xcosx + ∫ cosx dx
-xcosx + sinx + c

Question no. 9

If ∫x²lnx dx = x³lnx/m + x³/n + c , then what are the values of m and n respectively?

looks_one 1/3 , -1/9

looks_two 3,-9

looks_3 3,9

looks_4 3,3

Option looks_two 3,-9

Solution :
∫x²lnx dx
x³lnx/3 - ∫x³/3x dx
x³lnx/3 - x³/9 + c
m = 3 , n = -9

Question no. 10

What is ∫ 1/(1+e^x) dx equal to ?

looks_one x-logx +c

looks_two x-log(tanx) +c

looks_3 x- log(1+ e^x) + c

looks_4 log(1+ e^x) + c

Option looks_3 x- log(1+ e^x) + c

Solution :

Question no. 11

What is ∫ √x e^√x dx equal to ?

looks_one 2e^√x( x-2√x + 2 ) + c

looks_two 2e^√x( x + 2√x + 2 ) + c

looks_3 2e^√x( x + 2√x - 2 ) + c

looks_4 2e^√x( x-2√x - 2 ) + c

Option looks_one 2e^√x( x-2√x + 2 ) + c

Solution :
∫ √x e^√x dx
√x = t
dx = 2√xdt
∫ √x e^√x dx = 2 ∫ x e^t dt = 2 ∫ t² e^t dt
Integrate By parts
2 t² e^t - 2 ∫2te^t dt
2 t² e^t -4te^t + 4e^t + c
put t = √x
2e^√x( x-2√x + 2 ) + c

Question no. 12

What is ∫ secⁿxtanx dx equal to ?

looks_one (1/n)secⁿx + c

looks_two (1/(n-1))sec^n-1x + c

looks_3 (1/n)tanⁿx + c

looks_4 (1/(n-1))tan^n-1x + c

Option looks_one (1/n)secⁿx + c

Solution :
∫ secⁿxtanx dx
sec x = t
sec x tan x dx = dt
∫ sec^n-1x dt = ∫ t^n-1 dt
∫ t^n-1 dt = tⁿ/n + C = (1/n)secⁿx + C

Question no. 13

What is ∫ 1/(xlnx) dx equal to ?

looks_one ln(lnx) + c

looks_two lnx + c

looks_3 (lnx)² + c

looks_4 none of these

Option looks_one ln(lnx) + c

Solution :

Question no. 14

What is ∫ e^lnx dx equal to ?

looks_one xe^lnx + c

looks_two -xe^lnx + c

looks_3 x+ c

looks_4 x²/2 + c

Option looks_3 x+ c

Solution :
∫ e^lnx dx = ∫ 1 dx = x +c
∫ e^lnx dx = x + c

Question no. 15

What is ∫ lnx/x dx equal to ?

looks_one (lnx)²/2 + c

looks_two lnx/ 2 + c

looks_3 (lnx)² +c

looks_4 none of these

Option looks_one (lnx)²/2 + c

Solution :
∫ lnx/x dx
Integrate By parts

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