# INDEFINITE INTEGRALS USING PROPERTIES

Question 1 :

Integrate the following with respect to x:

e(tan x + log sec x)

Solution :

=  ∫e(tan x + log sec x) dx

Let f(x)  =  log sec x

Differentiating with respect to "x"

f'(x)  =  (1/sec x) (sec x tan x) dx

f'(x)  =  tan x

∫e(tan x + log sec x) dx  =   ∫e(f(x) + f'(x)) dx

=  ef(x) + c

=  elog sec x + c

Question 2 :

Integrate the following with respect to x:

e[(x - 1)/2x2]

Solution :

=  ∫e[(x - 1)/2x2] dx

=  ∫ex [(x/2x2) - (1/2x2)] dx

=  ∫e[(1/2x) - (1/2x2)] dx

Let f(x)  =  1/2x then f'(x)  =  -1/2x2

Hence ∫e[(1/2x) - (1/2x2)] dx  is in the form e(f(x) + f'(x)) dx

=  ef(x)

=  e(1/2x)

=  ex/2x

Question 3 :

Integrate the following with respect to x:

esec x (1 + tan x)

Solution :

=  ∫esec x (1 + tan x) dx

=  ∫ex (sec x  + sec x tan x) dx

Let f(x)  =  sec x, then f'(x)  =  sec x tan x

Hence e(sec x  + sec x tan x) dx is in the form e(f(x) + f'(x)) dx

=  ef(x)

=  esec x + c

Question 4 :

Integrate the following with respect to x:

e(2 + sin 2x) / (1 + cos 2x)

Solution :

=  ∫ex [(2 + sin 2x) / (1 + cos 2x)] dx

sin 2x  =  2 sin x cos x

1 + cos 2x  =  2 cos2x

=  ∫e[(2 + 2 sin x cos x) / 2 cos2x] dx

=  ∫e[(2 / 2 cos2x)  + (2 sin x cos x / 2 cos2x)] dx

=  ∫e[(1/cos2x)  + (sin x / cosx)] dx

=  ∫e[sec2x  + tan x] dx  ------(1)

Let f(x)  =  tan x then f'(x)  =  sec2x

Hence (1) is in the form

=  ∫e[f(x) + f'(x)] dx

=  ef(x) + c

=  etan x + c

Question 5 :

Integrate the following with respect to x:

e^tan-1x [(1 + x + x2) / (1 + x2)]

Solution :

In order to compare the given question with the form e[f(x) + f'(x)] dx, we must have power "x" to the power of "e".

Let t = tan-1x

x  =  tan t

dt  =  1/1 + x2

e^tan-1x [(1 + x + x2) / (1 + x2)] dx

=  ∫et (1 + tan t + tan2t) dt

Here 1 + tan2t  =  sec2t

=  ∫et (tan t + sec2t) dt

f(x)  =  tan t then f'(x)  =  sec2t

=  et tan t + c

=  e^tan-1x tan (tan-1x) + c

=  x e^tan-1x + c

Question 6 :

Integrate the following with respect to x:

log x / (1 + log x)2

Solution :

=  ∫ [log x / (1 + log x)2] dx

Let t = log x ==> et  =  x

Differentiating with respect to "x", we get

et dt =  dx

=  ∫ [t/(1 + t)2et dt

=  ∫ e[(t + 1 - 1) /(1 + t)2] dt

=  ∫ e[((1 + t)/(1 + t)2) - 1/(1 + t)2] dt

=  ∫ e[1/(1 + t) - 1/(1 + t)2] dt

If f(t)  =  1/(1 + t) then f'(t)  =  -1/(1 + t)2

=  ef(t) + c

= e (1/(1 + t)) + c

= x/(1 + log x) + c

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