**Indefinite Integral Using Properties :**

Here we are going to see some example problems to understand how to solve indefinite integral using properties.

**∫**e^{x}[ f(x) + f'(x) ] dx = e^{x} f (x) + c

**Question 1 :**

Integrate the following with respect to x:

e^{x }(tan x + log sec x)

**Solution :**

** = ∫****e ^{x }(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 ^{x }(tan x + log sec x) dx = **

** = ****e ^{x }f(x) + c **

** = ****e ^{x }log sec x + c **

**Question 2 :**

Integrate the following with respect to x:

e^{x }[(x - 1)/2x^{2}]

**Solution :**

** = ∫**e^{x }[(x - 1)/2x^{2}]** dx**

**= ∫**e^{x }[(x/2x^{2}) - (1/2x^{2})]** dx**

**= ∫**e^{x }[(1/2x) - (1/2x^{2})]** dx**

**Let f(x) = 1/2x then ****f'(x) = -1/2x ^{2}**

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

** = ****e ^{x }f(x)**

** = ****e ^{x }(1/2x)**

** = ****e**^{x}/2x

**Question 3 :**

Integrate the following with respect to x:

e^{x }sec x (1 + tan x)

**Solution :**

** = ∫**e^{x }sec x (1 + tan x) dx

** = ∫**e^{x }(sec x + sec x tan x) dx

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

**Hence ****∫**e^{x }(sec x + sec x tan x) dx** is in the form ****∫****e ^{x }(f(x) + f'(x)) dx**

** = ****e ^{x }f(x)**

** = ****e ^{x }sec x**

**Question 4 :**

Integrate the following with respect to x:

e^{x }(2 + sin 2x) / (1 + cos 2x)

**Solution :**

** = ∫**e^{x }[(2 + sin 2x) / (1 + cos 2x)] dx

sin 2x = 2 sin x cos x

1 + cos 2x = 2 cos^{2}x

** = ∫**e^{x }[(2 + 2 sin x cos x) / 2 cos^{2}x] dx

** = ∫**e^{x }[(2 / 2 cos^{2}x) + (2 sin x cos x / 2 cos^{2}x)] dx

** = ∫**e^{x }[(1/cos^{2}x) + (sin x / cosx)] dx

** = ∫**e^{x }[sec^{2}x + tan x] dx ------(1)

Let f(x) = tan x then f'(x) = sec^{2}x

Hence (1) is in the form

** = ∫**e^{x }[f(x) + f'(x)] dx

= e^{x }f(x) + c

= e^{x }tan x + c

**Question 5 :**

Integrate the following with respect to x:

e^tan^{-1}x [(1 + x + x^{2}) / (1 + x^{2})]

**Solution :**

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

Let t = tan^{-1}x

x = tan t

dt = 1/1 + x^{2}

**∫**e^tan^{-1}x [(1 + x + x^{2}) / (1 + x^{2})] dx

= **∫e ^{t} (1 + tan t + tan^{2}t) dt**

**Here 1 + ****tan ^{2}t = sec^{2}t**

** = ∫e^{t} (tan t + sec^{2}t) dt**

**f(x) = tan t then f'(x) = ****sec ^{2}t**

**= ****e ^{t }tan t + c**

**= e^**tan

**= x e^**tan

**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 ==> e^{t} = x

Differentiating with respect to "x", we get

e^{t} dt = dx

= ∫ [t/(1 + t)^{2}] e^{t} dt

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

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

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

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

= e^{t }f(t) + c

= e^{t } (1/(1 + t)) + c

= x/(1 + log x) + c

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