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Indefinite integration

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Question no. 16

What is ∫ sin^-1x + cos^-1x dx is equal to ?

looks_one 0.5πx + c

looks_two x(sin^-1x + cos^-1x) + c

looks_3 x(sin^-1x - cos^-1x) + c

looks_3 π/2 + x + c

Option looks_one 0.5πx + c

Solution :
∫ sin^-1x + cos^-1x dx
∫ π/2 dx
0.5πx + c

Question no. 17

What is ∫ (x-1)/(x+1)² dx equal to ?

looks_one log(x+1) + 2/(x+1) +c

looks_two log(x+1) - 2/(x+1) +c

looks_3 -log(x+1) + 2/(x+1) +c

looks_4 none of these

Option looks_one log(x+1) + 2/(x+1) +c

Solution :
∫ (x-1)/(x+1)² dx
x + 1 = t ⇒ x = t -1 ⇒ dx = dt
∫ (t-2)/t² dt
∫ 1/t - 2/t² dt
log t + 2/t + c
log(x+1) + 2/(x+1) +c

Question no. 18

What is ∫ (1+ 2x +3x² +4x³...........) dx equal to ?

looks_one (1+x)^-1 + c

looks_two (1-x)^-1 + c

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

looks_4 none of these

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

Solution :
(1+ 2x +3x² +4x³...........) = (1-x)^-2
∫ (1+ 2x +3x² +4x³...........) dx = ∫ (1-x)^-2 dx
∫ (1-x)^-2 = (1-x)^-1 + c

Question no. 19

What is ∫ sin^-1(cosx)dx equal to ?

looks_one xπ/2 - x²/2 + k

looks_two xπ/2 + x²/2 + k

looks_3 -xπ/2 - x²/2 + k

looks_4 π/2 - x²/2 + k

Option looks_one xπ/2 - x²/2 + k

Solution :
∫ sin^-1(cosx)dx
∫ sin^-1(sin (π/2 - x))dx
∫ π/2 - x dx
xπ/2 - x²/2 + k

Question no. 20

What is ∫ sin√x /√x dx equal to ?

looks_one 0.5cos√x + c

looks_two cos√x + c

looks_3 -0.5cos√x + c

looks_4 2cos√x + c

Option looks_4 2cos√x + c

Solution :
∫ sin√x /√x dx
√x = t
dx = 2√xdt
2 ∫ sint dt
2 cos t + c
2cos√x + c

Question no. 21

What is ∫ e^xsinx dx equal to ?

looks_one e^x(sinx +cosx) + c

looks_two e^x(sinx - cosx) + c

looks_3 0.5e^x(sinx +cosx) + c

looks_4 0.5e^x(sinx -cosx) + c

Option looks_4 0.5e^x(sinx -cosx) + c

Solution :
∫ e^xsinx dx
Integrate By parts
∫ e^xsinx dx = e^xsinx - ∫ e^xcosx dx
∫ e^xsinx dx = e^xsinx - e^xcosx - ∫ e^xsinx dx
2 ∫ e^xsinx dx = e^xsinx - e^xcosx
∫ e^xsinx dx = 0.5e^x(sinx -cosx) + c

Question no. 22

What is ∫ x sinkx dx equal to ?

looks_one (1/k)sinkx + c

looks_two (1/k)coskx +c

looks_3 (1/k)sinx + c

looks_4 none of these

Option looks_4 none of these

Solution :
∫ x sinkx dx
kx = t ⇒ dx = dt/k
(1/k²) ∫ t sin t dt dt
(1/k²) [-tcost + sint ] = (1/k²)[-kxcoskt + sinkt]

Question no. 23

What is ∫ xe^x dx equal to ?

looks_one (x+1)e^x + c

looks_two xe^x - 1+ c

looks_3 xe^x + 1+ c

looks_4 (x-1)e^x + c

Option looks_4 (x-1)e^x + c

Solution :
∫ xe^x dx
xe^x - ∫ e^x dx
xe^x - e^x + c
(x-1)e^x + c

Question no. 24

If ∫xe^2x dx = e^2xf(x) + c , then f(x) is

looks_one (3x-1)/4

looks_two (2x+1)/2

looks_3 (2x-1)/4

looks_4 (x-4)/6

Option looks_3 (2x-1)/4

Solution :
∫xe^2x dx
Integrate by parts
(1/2)xe^2x - (1/2)∫e^2x dx
(1/2)xe^2x - (1/4)e^2x + C
(2x-1)e^2x/4 + C
f(x) = (2x-1)/4

Question no. 25

What is ∫ 1/(x+1)(x+2) dx equal to ?

looks_one log(x+2)/(x+1) +c

looks_two log(x+2)(x+1) +c

looks_3 log(x+1)/(x+2) +c

looks_4 none of these

Option looks_3 log(x+1)/(x+2) +c

Solution :
∫ 1/(x+1)(x+2) dx
∫ 1/(x+1) - 1/(x+2) dx
log(x+1) - log(x+2) + c
log(x+1)/(x+2) +c

Question no. 26

What is ∫ 1/(x-x³) dx equal to ?

looks_one 0.5log(1- x²)/x² + c

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

looks_3 logx(1- x²) + c

looks_4 0.5log x²/(1-x²) + c

option looks_4 0.5log x²/(1-x²) + c

Solution :

Question no. 27

What is ∫ 1/(e^x +e^-x)² dx equal to ?

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

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

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

looks_4 none of these

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

Solution :

Question no. 28

What is ∫ cos³x e^logsinx dx equal to ?

looks_one -(1/4)sin⁴x + c

looks_two (1/4)cos⁴x + c

looks_3 (1/4)e^sinx + c

looks_4 none of these

Option looks_4 none of these

Solution :
∫ cos³x e^logsinx dx
∫ cos³x sinx dx
sinx = t ⇒ cosx dx = dt
∫ cos²x t dt
∫ (1 -t²)t dt
∫ t - t³ dt = t²/2 - t⁴/4 + c
(1/2)sin²x - (1/4)sin⁴x + c

Question no. 29

What is ∫ tan(3x-5)sec(3x-5) dx equal to ?

looks_one sec(3x-5) + c

looks_two (1/3)sec(3x-5) + c

looks_3 tan(3x-5)+ c

looks_4 none of these

Option looks_two (1/3)sec(3x-5) + c

Solution :
∫ tan(3x-5)sec(3x-5) dx
3x-5 = t ⇒ 3dx = dt
(1/3) ∫ tan(t)sec(t) dt
(1/3){sect} + c
(1/3)sec(3x-5) + c

Question no. 30

What is ∫ (1 + x + x²/2! + x³/3! + ...........)dx equal to ?

looks_one -e^x + c

looks_two e^x + c

looks_3 e^-x +c

looks_4 none of these

Option looks_two e^x + c

Solution :
e^x = 1 + x + x²/2! + x³/3! + ...........
∫ (1 + x + x²/2! + x³/3! + ...........) = ∫ e^x dx
∫ e^x dx = e^x + c

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