Evaluate the following integral : ∫ x 5 x - 1 d x

Given, ∫ x 5 x - 1 d x = ∫ x 4 x x - 1 d x = ∫ x 4 x - 1 + 1 x - 1 d x = ∫ x 4 + x 4 x - 1 d x = ∫ x 4 + x 3 x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 3 x - 1 d x = ∫ x 4 + x 3 + x 2 x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x 2 x - 1 d x = ∫ x 4 + x 3 + x 2 + x x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x + x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x + 1 + 1 x - 1 d x ∴ ∫ x 5 x - 1 d x = x 5 5 + x 4 4 + x 3 3 + x 2 2 + x + log x - 1 + c .

Evaluate the following integral : ∫ x 5 x - 1 d x

Given, ∫ x 5 x - 1 d x = ∫ x 4 x x - 1 d x = ∫ x 4 x - 1 + 1 x - 1 d x = ∫ x 4 + x 4 x - 1 d x = ∫ x 4 + x 3 x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 3 x - 1 d x = ∫ x 4 + x 3 + x 2 x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x 2 x - 1 d x = ∫ x 4 + x 3 + x 2 + x x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x + x - 1 + 1 x - 1 d x = ∫ x 4 + x 3 + x 2 + x + 1 + 1 x - 1 d x ∴ ∫ x 5 x - 1 d x = x 5 5 + x 4 4 + x 3 3 + x 2 2 + x + log x - 1 + c .

Evaluate the following integral : ∫ d x e x + 1

Given, ∫ d x e x + 1 = ∫ 1 - e x e x + 1 d x = ∫ d x - ∫ e x e x + 1 d x Let t = e x + 1 Differentiating both sides, we get ⇒ d t = e x d x So, ∫ d x - ∫ e x e x + 1 d x = ∫ d x - ∫ 1 t d t = x - log t + c ∴ ∫ d x e x + 1 = x - log e x + 1 + c .

Evaluate the following integral : ∫ x 2 + 1 x 4 d x

Given, ∫ x 2 + 1 x 4 d x = ∫ x 1 + 1 x 2 x 4 d x = ∫ 1 + 1 x 2 x 3 d x Let t = 1 + 1 x 2 Differentiating both sides w.r.t x , we get d t d x = - 2 x 3 ⇒ d t - 2 = 1 x 3 d x So, ∫ 1 + 1 x 2 x 3 d x = - 1 2 ∫ t d t = - 1 3 t 3 2 + c ∴ ∫ x 2 + 1 x 4 d x = - 1 3 1 + 1 x 2 3 2 + c .

Evaluate the following integral : ∫ tan x cos 4 x d x

Given, ∫ tan x cos 4 x d x = ∫ tan x sec 4 x d x = ∫ tan x sec 2 x sec 2 x d x = ∫ tan x tan 2 x + 1 sec 2 x d x ∵ sec 2 x - tan 2 x = 1 = ∫ tan 5 2 x + tan x sec 2 x d x Let t = tan x Differentiating both sides w.r.t x , we get d t = sec 2 x d x So, = ∫ tan 5 2 x + tan x sec 2 x d x = ∫ t 5 2 + t 1 2 d t = t 7 2 7 2 + t 3 2 3 2 + c = 2 t 7 2 7 + 2 t 3 2 3 + c ∴ ∫ tan x cos 4 x d x = 2 tan x 7 2 7 + 2 tan x 3 2 3 + c .

Evaluate the following integral : ∫ x x 1 + log x d x

Given, ∫ x x 1 + log x d x = ∫ e x · log x 1 + log x d x Let t = x x ⇒ e x · log x = t Differentiating both sides w.r.t x , we get e x · log x · x · 1 x + log x d x = d t ⇒ e x · log x · 1 + log x d x = d t Now, ∫ x x 1 + log x d x = ∫ d t = t + c = e x · log x + c ∴ ∫ x x 1 + log x d x = x x + c .

Evaluate the following integral : ∫ e x x x · log x + 1 d x

Given, ∫ e x x x · log x + 1 d x = ∫ e x log x + 1 x d x Let z = e x · log x Differentiating both sides w.r.t x , we get e x · 1 x + e x · log x = d z d x ⇒ e x · 1 x + log x d x = d z Now, ∫ e x 1 x + log x d x = ∫ d z = z + c = e x · log x + c ∴ ∫ e x x x · log x + 1 d x = e x · log x + c .

Evaluate the following integral : ∫ x e x x + 1 2 d x

Given, ∫ x e x x + 1 2 d x Let e x 1 + x = t Differentiating both sides w.r.t x , we get x + 1 e x - e x x + 1 2 = d t d x ⇒ e x x + 1 2 x + 1 - 1 d x = d t Now, ∫ x e x x + 1 2 d x = ∫ d t = t + c = e x x + 1 + c ∴ ∫ x e x x + 1 2 d x = e x x + 1 + c .

Evaluate the following integral : ∫ e x x - 2 x 3 d x

Given, ∫ e x x - 2 x 3 d x = ∫ e x x x 3 - 2 x 3 d x = ∫ e x 1 x 2 - 2 x 3 d x Let z = e x x 2 Differentiating both sides w.r.t x , we get e x · 1 x 2 + e x - 2 x 3 = d z d x ⇒ e x · 1 x 2 - 2 x 3 d x = d z Now, ∫ e x x - 2 x 3 d x = ∫ d z = z + c = e x x 2 + c ∴ ∫ e x x - 2 x 3 d x = e x · x - 2 + c .

Evaluate the following integral : ∫ e x 1 - x 2 1 + x 2 2 d x

Given, ∫ e x 1 - x 2 1 + x 2 2 d x = ∫ e x 1 + x 2 - 2 x 1 + x 2 2 d x = ∫ e x 1 1 + x 2 - 2 x 1 + x 2 2 d x Let z = e x 1 + x 2 Differentiating both sides w.r.t x , we get e x · 1 1 + x 2 + e x - 1 1 + x 2 2 · 2 x = d z d x ⇒ e x · 1 1 + x 2 - 2 x 1 + x 2 2 d x = d z Now, ∫ e x 1 - x 2 1 + x 2 2 d x = ∫ d z = z + c = e x 1 + x 2 + c ∴ ∫ e x 1 - x 2 1 + x 2 2 d x = e x 1 + x 2 + c .