Evaluate ∫ x + x 2 + 2 d x .

Let us assume I = ∫ x + x 2 + 2 d x Now put x + x 2 + 2 = t x 2 + 2 = t 2 + x 2 − 2 t x ⇒ x = t 2 − 2 2 t Differentiate both sides w.r.t. t d x d t = t 2 + 2 2 t 2 Or, d x = t 2 + 2 d t 2 t 2 Replacing this in I , we get I = ∫ t 1 2 t 2 + 2 2 t 2 d t ⇒ I = ∫ t 1 2 · t 2 + 2 t 1 2 2 t 2 ⇒ I = 1 2 ∫ t 1 2 d t + ∫ t - 3 2 d t ⇒ I = 1 2 t 1 2 + 1 1 2 + 1 + t - 3 2 + 1 - 3 2 + 1 + C ⇒ I = 1 3 t 3 2 - 2 t - 1 2 + C Now, replacing the value of t I = 1 3 x + x 2 + 2 3 2 − 2 1 x + x 2 + 2 + C

Evaluate ∫ x + x 2 + 2 d x .

Let us assume I = ∫ x + x 2 + 2 d x Now put x + x 2 + 2 = t x 2 + 2 = t 2 + x 2 − 2 t x ⇒ x = t 2 − 2 2 t Differentiate both sides w.r.t. t d x d t = t 2 + 2 2 t 2 Or, d x = t 2 + 2 d t 2 t 2 Replacing this in I , we get I = ∫ t 1 2 t 2 + 2 2 t 2 d t ⇒ I = ∫ t 1 2 · t 2 + 2 t 1 2 2 t 2 ⇒ I = 1 2 ∫ t 1 2 d t + ∫ t - 3 2 d t ⇒ I = 1 2 t 1 2 + 1 1 2 + 1 + t - 3 2 + 1 - 3 2 + 1 + C ⇒ I = 1 3 t 3 2 - 2 t - 1 2 + C Now, replacing the value of t I = 1 3 x + x 2 + 2 3 2 − 2 1 x + x 2 + 2 + C

Evaluate: ∫ d x x 3 + 3 x 2 + 3 x + 1 x 2 + 2 x - 3

Let the given integral ∫ d x x 3 + 3 x 2 + 3 x + 1 x 2 + 2 x - 3 = I ⇒ I = ∫ d x x + 1 3 x + 1 2 - 4 Let x + 1 = t ⇒ d x = d t ⇒ I = ∫ d t t 3 t 2 - 4 Again consider t 2 - 4 = u ⇒ t 2 = u 2 + 4 ⇒ t t 2 - 4 d t = d u ⇒ d t = t 2 - 4 t d u ⇒ I = ∫ d u u 2 + 4 2 Let u = 2 tan θ ⇒ d u = 2 sec 2 θ d θ ⇒ I = ∫ 2 sec 2 θ 4 + 4 tan 2 θ 2 d θ = 1 8 ∫ cos 2 θ d θ = 1 8 ∫ 1 + cos 2 θ 2 d θ = 1 16 ∫ d θ + ∫ cos 2 θ d θ = 1 16 θ + sin 2 θ 2 + C , where C is arbitrary integrating constant. ⇒ I = 1 16 θ + 1 32 2 tan θ 1 + tan 2 θ + C = 1 16 tan - 1 u 2 + 1 16 u 2 1 + u 2 2 + C = 1 16 cos - 1 2 u 2 + 4 + 1 8 u u 2 + 4 + C = 1 16 cos - 1 2 t 2 - 4 + 4 + 1 8 t 2 - 4 t 2 - 4 + 4 + C = 1 16 cos - 1 2 t + 1 8 x + 1 2 - 4 x + 1 2 + C ⇒ I = x 2 + 2 x - 3 8 ( x + 1 ) 2 + 1 16 · cos - 1 2 x + 1 + C

Evaluate : ∫ ( cos 2 x - 3 ) cos 4 x 4 - cot 2 x d x

Consider I = ∫ cos 2 x - 3 cos 4 x 4 - cot 2 x d x ⇒ I = ∫ 1 - tan 2 x 1 + tan 2 x - 3 sec 4 x 4 - 1 tan 2 x d x = ∫ tan x - 4 tan 2 x - 2 4 tan 2 x - 1 sec 2 x d x Let tan x = t ⇒ sec 2 x d x = d t ⇒ I = ∫ t - 4 t 2 - 2 4 t 2 - 1 d t = - 2 ∫ t 2 t 2 + 1 4 t 2 - 1 d t Let 4 t 2 - 1 = u 2 ⇒ 8 t d t = 2 u d u ⇒ 2 t d t = u 2 d u ⇒ I = - ∫ 2 u 2 + 1 4 + 1 u 2 × u 2 d u = - 1 2 ∫ u 2 2 + 3 2 d u = - 1 4 u 3 3 + 3 u + C , where C is arbitrary integrating constant. = - 1 4 4 t 2 - 1 3 / 2 3 + 3 4 t 2 - 1 1 / 2 + C = - 1 4 4 t 2 - 1 4 t 2 - 1 3 + 3 + C = - 1 3 4 tan 2 x - 1 tan 2 x + 2 + C ∴ I = C - 1 3 tan x · 2 + tan 2 x · 4 - cot 2 x

Evaluate : ∫ x 2 + 1 ln x 2 + 1 - 2 ln x x 4 d x

Let I = ∫ x 2 + 1 ln x 2 + 1 - 2 ln x x 4 d x = ∫ x 2 + 1 ln x 2 + 1 - ln x 2 x 4 d x = ∫ 1 + 1 x 2 × ln x 2 + 1 x 2 x 3 d x = ∫ 1 + 1 x 2 × ln 1 + 1 x 2 x 3 d x Let 1 + 1 x 2 = t ⇒ - 2 x 3 d x = d t ⇒ I = - 1 2 ∫ t ln t d t By integration by parts and consider first function is ln t and second function is t , we get ⇒ I = - 1 2 ln t ∫ t 1 / 2 d t - ∫ d d t ln t ∫ t 1 / 2 d t d t = - 1 2 2 3 ln t · t 3 / 2 - 2 3 ∫ t 3 / 2 t d t + C , where C is arbitrary integrating constant. = - 1 2 2 3 ln t · t 3 / 2 - 4 9 t 3 / 2 + C = 1 9 t 3 / 2 2 - 3 ln t + C = 2 9 1 + 1 x 2 3 / 2 1 - 3 2 ln 1 + 1 x 2 + C ∴ I = 2 x 2 + 1 x 2 + 1 9 x 3 · 1 - 3 2 ln 1 + 1 x 2 + C

Evaluate : ∫ x 7 x - 10 - x 2 3 / 2 d x

Consider ∫ x 7 x - 10 - x 2 3 / 2 d x = I ⇒ I = ∫ x - x 2 - 7 x + 10 3 / 2 d x = ∫ x - x 2 - 2 x - 5 x + 10 3 / 2 d x = ∫ x 2 - x x - 5 3 / 2 d x Let x = 5 cos 2 θ + 2 sin 2 θ ⇒ d x = - 10 cos θ sin θ + 4 sin θ cos θ d θ = - 6 sin θ cos θ d θ ⇒ I = ∫ 5 cos 2 θ + 2 sin 2 θ - 3 cos 2 θ - 3 sin 2 θ 3 / 2 - 6 sin θ cos θ d θ = - 6 27 ∫ 5 cos 2 θ + 2 sin 2 θ sin θ cos θ 2 d θ = - 6 27 ∫ 5 cosec 2 θ d θ + ∫ 2 sec 2 θ d θ = - 2 9 - 5 cot θ + 2 tan θ + C , where C is arbitrary constant. = 10 9 cot θ - 4 9 tan θ + C Since x = 5 cos 2 θ + 2 sin 2 θ = 5 1 - sin 2 θ + 2 sin 2 θ = 5 - 3 sin 2 θ ⇒ sin θ = 5 - x 3 ⇒ cos θ = 1 - sin 2 θ = 1 - 5 - x 3 = x - 2 3 Hence I = 10 9 x - 2 5 - x - 4 9 5 - x x - 2 + C ⇒ I = 10 x - 2 - 4 5 - x 9 5 - x x - 2 + C ∴ I = 2 ( 7 x - 20 ) 9 7 x - 10 - x 2 + C

If ∫ x cos α + 1 x 2 + 2 x cos α + 1 3 2 d x = f x g x + C , then find f x and g x .

Consider I = ∫ x cos α + 1 x 2 + 2 x cos α + 1 3 2 d x ⇒ I = ∫ x cos α + 1 x 3 1 + 2 x cos α + 1 x 2 3 2 d x ⇒ I = ∫ cos α x 2 + 1 x 3 1 + 2 x cos α + 1 x 2 3 2 d x Put 1 + 2 x cos α + 1 x 2 = t ⇒ - 2 x 2 cos α + - 2 x 3 d x = d t ⇒ I = - 1 2 ∫ d t t 3 / 2 = - 1 2 t - 3 2 + 1 - 3 2 + 1 + C , where C is arbitrary constant. ⇒ I = 1 t + C = 1 1 + 2 x cos α + 1 x 2 + C = x x 2 + 2 x cos α + 1 + C … 1 Also given that ∫ x cos α + 1 x 2 + 2 x cos α + 1 3 2 d x = f x g x + C … 2 Comparing equations 1 & 2 , we get f x = x and g x = x 2 + 2 x cos α + 1

Evaluate: ∫ e x x 3 - x + 2 x 2 + 1 2 d x

Consider the given integral ∫ e x x 3 - x + 2 x 2 + 1 2 d x = I ⇒ I = ∫ e x x 3 + x - 2 x - x 2 + 1 + 1 + x 2 x 2 + 1 2 d x = ∫ e x x 3 + x x 2 + 1 2 + - 2 x x 2 + 1 2 + 1 - x 2 x 2 + 1 2 + 1 + x 2 x 2 + 1 2 d x = ∫ e x x x 2 + 1 + 1 1 + x 2 + 1 - x 2 x 2 + 1 2 + - 2 x x 2 + 1 2 d x = ∫ e x 1 + x x 2 + 1 + 1 - x 2 - 2 x x 2 + 1 2 d x From integration by parts we know that ∫ e x f x + f ' x d x = e x f x + C . Here I = ∫ e x 1 + x x 2 + 1 + d d x 1 + x x 2 + 1 d x = e x 1 + x x 2 + 1 + C , where C is arbitrary integrating constant. ⇒ ∫ e x x 3 - x + 2 x 2 + 1 2 d x = e x x + 1 x 2 + 1 + C

Evaluate: ∫ x 2 x sin x + cos x 2 d x

Consider I = ∫ x 2 x sin x + cos x 2 d x ⇒ I = ∫ x cos x · x cos x x sin x + cos x 2 d x Apply integration by parts and assume first function is x cos x and second function is x cos x x sin x + cos x 2 , we get I = x cos x ∫ x cos x x sin x + cos x 2 d x - ∫ d d x x cos x ∫ x cos x x sin x + cos x 2 d x d x ⇒ I = x cos x ∫ x cos x x sin x + cos x 2 d x - ∫ cos x + x sin x cos 2 x ∫ x cos x x sin x + cos x 2 d x d x … 1 Let I 1 = ∫ x cos x x sin x + cos x 2 d x Substitute x sin x + cos x = t ⇒ x cos x + sin x - sin x d x = d t ⇒ x cos x d x = d t So, I 1 = ∫ d t t 2 = - 1 t = - 1 x sin x + cos x Put this value in equation 1 , we get I = x cos x - 1 x sin x + cos x - ∫ x sin x + cos x cos 2 x · - 1 x sin x + cos x d x = - x cos x x sin x + cos x + ∫ sec 2 x d x = - x cos x x sin x + cos x + tan x + C , where C is arbitrary constant. = - x cos x x sin x + cos x + sin x cos x + C = 1 cos x - x x sin x + cos x + sin x + C = 1 cos x cos x sin x - x 1 - sin 2 x x sin x + cos x + C = 1 cos x sin x cos x - x cos 2 x x sin x + cos x + C ⇒ I = sin x - x cos x x sin x + cos x + C

Evaluate: ∫ sin 4 x · e tan 2 x d x

Consider the given integral ∫ sin 4 x · e tan 2 x d x = I Since sin 2 x = 2 sin x cos x ⇒ I = ∫ 2 sin 2 x cos 2 x e tan 2 x d x = ∫ 2 2 tan x 1 + tan 2 x 1 - tan 2 x 1 + tan 2 x e tan 2 x d x ⇒ I = ∫ 4 tan x 1 - tan 2 x sec 4 x e tan 2 x d x Let tan 2 x = t ⇒ 2 tan x sec 2 x d x = d t And sec 2 x = 1 + tan 2 x = 1 + t ⇒ I = ∫ 2 1 - t 1 + t 3 e t d t = 2 ∫ 2 - 1 + t 1 + t 3 e t d t = 2 ∫ e t 2 1 + t 3 - 1 + t 1 + t 3 d t = 2 ∫ e t 2 1 + t 3 - 1 1 + t 2 d t = 2 ∫ e t - 1 1 + t 2 + 2 1 + t 3 d t Since ∫ e x f x + f ' x d x = e x f x + C ⇒ I = 2 ∫ e t - 1 1 + t 2 + d d t - 1 1 + t 2 d t = 2 e t - 1 1 + t 2 + C = - 2 e tan 2 x 1 + tan 2 x 2 + C = - 2 e tan 2 x sec 2 x 2 + C ∴ I = - 2 cos 4 x · e tan 2 x + C