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QUESTION: 1

One value of ∫ f'(x) dx is

Solution:

As (∫f'(x)dx) = f(x) + C, therefore, one value of (∫ f'(x) dx) is f(x)

QUESTION: 2

∫ log x dx is equal to

Solution:

QUESTION: 3

Solution:

QUESTION: 4

Solution:

**ANSWER :- a**

**Solution :- Let e^x/2 = tanθ. Then 1/2e^(x/2)dx = sec^2θdθ.**

**∫dx/(e^x+1) = 2∫½ e^(x/2)dx/[e^(x/2)(e^x+1)]**

**= 2∫[sec^2θdθ]/tanθ(sec^2θ)**

**= 2∫cosθ/sinθ)dθ = 2ln |sinθ|**

**From tanθ = e^(x/2) draw a right triangle to see that sinθ = e^x/2√e^x+1:**

**= 2ln∣e^pi/2√(e^x + 1)|**

**= ln |e^x/(e^x+1)|**

**= x−ln(e^x+1)+ C**

QUESTION: 5

Solution:

QUESTION: 6

if ∫ g(x) dx = g(x), then is equal to

Solution:

QUESTION: 7

If f (x) be a function such that

Solution:

∫ log xdx = xlog x - x = x(log x -1)

= x(log x - log e) =

QUESTION: 8

Solution:

QUESTION: 9

Solution:

Dividing numerator and denominator by cos2x and substituting tanx = t , we get :

QUESTION: 10

Solution:

QUESTION: 11

∫ sec^{2} x cosec^{2} x dx is equal to

Solution:

QUESTION: 12

Solution:

QUESTION: 13

Solution:

QUESTION: 14

Solution:

QUESTION: 15

Solution:

Note that sin^{9}x is an odd function, therefore,

QUESTION: 16

Solution:

QUESTION: 17

Solution:

∫ log(1-x) dx - ∫ log x dx = 0

QUESTION: 18

Solution:

QUESTION: 19

Solution:

QUESTION: 20

Solution:

QUESTION: 21

∫ (tan x + cot x) dx is equal to

Solution:

**Correct Answer :- c**

**Explanation : ∫(tanx + cotx)dx**

**= ∫tanxdx + ∫cotxdx**

**= ∫sinx/cosx dx + ∫cosx/sinx dx**

**I1 = sinx/cosx**

**Put t = cosx**

**dt = -sinx dx**

**I2 = cosx/sinx**

**Put t = sinx**

**dt = cosx dx**

**So, we get**

**=> ∫-dt/t + ∫dt/t**

**=> - ln t + ln t + c**

**=> -ln cosx + ln sinx + c**

**=> log (sinx/cosx) + c**

**=> log(tanx)**

QUESTION: 22

∫ log(log x) + (log x)^{-1}) dx is equal to

Solution:

QUESTION: 23

Solution:

QUESTION: 24

Solution:

QUESTION: 25

Solution:

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