Find ∫ 2 x + 4 x 2 + 4 x + 3 d x

Let I = ∫ 2 x + 4 x 2 + 4 x + 3 d x Let x 2 + 4 x + 3 = t ⇒ 2 x + 4 d x = d t ∴ I = ∫ t d t = t 3 2 3 2 + C = 2 3 t 3 2 + C = 2 3 x 2 + 4 x + 3 3 2 + C , where C is constant of integration.

Find ∫ 2 x + 4 x 2 + 4 x + 3 d x

Let I = ∫ 2 x + 4 x 2 + 4 x + 3 d x Let x 2 + 4 x + 3 = t ⇒ 2 x + 4 d x = d t ∴ I = ∫ t d t = t 3 2 3 2 + C = 2 3 t 3 2 + C = 2 3 x 2 + 4 x + 3 3 2 + C , where C is constant of integration.

Find ∫ 1 + x 2 x 2 e x - 1 x d x

Let I = ∫ 1 + x 2 x 2 e x - 1 x d x Let x - 1 x = t ⇒ 1 - - 1 x 2 d x = d t ⇒ x 2 + 1 x 2 d x = d t ∴ I = ∫ e t d t = e t + C = e x - 1 x + C , where C is constant of integration.

Let I = ∫ d x x + x = ∫ d x x x + 1 Let x + 1 = t ⇒ 1 2 x d x = d t ⇒ d x x = 2 d t ∴ I = ∫ 2 d t t = 2 log e t + C = 2 log e x + 1 + C , where C is constant of integration.

Find ∫ tan 2 x sec 2 x 1 + tan 6 x d x .

Let I = ∫ tan 2 x sec 2 x 1 + tan 6 x d x Let tan 3 x = t ⇒ 3 tan 2 x · sec 2 x d x = d t ⇒ tan 2 x · sec 2 x d x = d t 3 ∴ I = 1 3 ∫ d t 1 + t 2 = 1 3 tan - 1 t + C = 1 3 tan - 1 tan 3 x + C , where C is constant of integration.

Find ∫ d x x log e x log e log e x

Let I = ∫ d x x log e x log e log e x Let log e log e x = t ⇒ 1 log e x × 1 x d x = d t ∴ I = ∫ d t t = log e t + C = log e log e log e x + C , where C is constant of integration.

Find ∫ e 2 x 4 e x + 1 d x

Let I = ∫ e 2 x 4 e x + 1 d x = ∫ e x · e x d x 4 e x + 1 Let e x + 1 = t ⇒ e x d x = d t ∴ I = ∫ t - 1 d t 4 t = 1 4 ∫ t - t - 1 2 d t = 1 4 t 1 2 + 1 1 2 + 1 + t - 1 2 + 1 - 1 2 + 1 + C = 1 2 t 3 2 3 - t 1 2 + C = t 1 2 6 t - 3 + C = e x + 1 6 e x + 1 - 3 + C = e x + 1 6 e x - 2 + C , where C is constant of integration.

Find ∫ d x x x 2 n - a 2 n

Let I = ∫ d x x x 2 n - a 2 n Multiply and divide by x n - 1 . ⇒ ∫ x n - 1 d x x n x n 2 - a n 2 Let x n = t ⇒ n x n - 1 d x = d t ⇒ x n - 1 d x = d t n ∴ I = 1 n ∫ d t t t 2 - a n 2 = 1 n · 1 a n sec - 1 t a n + C = 1 n a - n sec - 1 x a n + C , where C is constant of integration.

Find ∫ sec x + 1 d x

I = ∫ sec x + 1 d x = ∫ 1 + cos x cos x d x = ∫ 2 cos 2 x 2 1 - 2 sin 2 x 2 d x = ∫ 2 cos x 2 1 - 2 sin 2 x 2 d x Case I: cos x 2 ≥ 0 I = ∫ 2 cos x 2 1 - 2 sin 2 x 2 d x Let 2 sin x 2 = t ⇒ 2 · 1 2 cos x 2 d x = d t ⇒ cos x 2 d x = 2 d t ∴ I = ∫ 2 · 2 d t 1 - t 2 = 2 sin - 1 t + C = 2 sin - 1 2 sin x 2 + C , where C is constant of integration. Case II: cos x 2 < 0 I = ∫ - 2 cos x 2 1 - 2 sin 2 x 2 d x Let 2 sin x 2 = t ⇒ 2 · 1 2 cos x 2 d x = d t ⇒ cos x 2 d x = 2 d t ∴ I = - ∫ 2 · 2 d t 1 - t 2 = - 2 sin - 1 t + C = - 2 sin - 1 2 sin x 2 + C , where C is constant of integration.

Find ∫ sin 2 x sin 4 x + cos 4 x d x .

I = ∫ sin 2 x sin 4 x + cos 4 x d x = ∫ sin 2 x sin 2 x 2 + cos 2 x 2 d x = ∫ sin 2 x sin 2 x + cos 2 x 2 - 2 sin 2 x cos 2 x d x = ∫ sin 2 x 1 - 2 sin 2 x cos 2 x d x = ∫ 2 sin 2 x 2 - 4 sin 2 x cos 2 x d x = ∫ 2 sin 2 x 1 + 1 - sin 2 2 x { ∵ sin 2 x = 2 sin x cos x } = ∫ 2 sin 2 x 1 + cos 2 2 x { ∵ cos 2 x = 1 - sin 2 x } Let cos 2 x = t ⇒ - 2 sin 2 x d x = d t ⇒ 2 sin 2 x d x = - d t ∴ I = ∫ - d t 1 + t 2 = - tan - 1 t + C = - tan - 1 cos 2 x + C , where C is constant of integration.