Consider I = ∫ 2 x 3 e x 2 d x ⇒ I = ∫ 2 x · x 2 e x 2 d x Let x 2 = t ⇒ 2 x d x = d t ⇒ I = ∫ t e t d t Apply integration by parts by considering first function is t and second function is e t , we get I = t ∫ e t d t - ∫ d t d t ∫ e t d t d t ⇒ I = t e t - e t + C , where C is arbitrary constant. So, I = e x 2 x 2 - 1 + C .

Given I = ∫ sin - 1 x d x Let sin - 1 x = t ⇒ x = sin 2 t ⇒ d x = 2 sin t cos t d t = sin 2 t d t So, I = ∫ t sin 2 t d t Apply integration by parts by considering first function is t and second function is sin 2 t , we get I = t ∫ sin 2 t d t - ∫ d t d t ∫ sin 2 t d t d t = - t cos 2 t 2 + sin 2 t 4 + C , where C is arbitrary constant. = - t 2 1 - 2 sin 2 t + sin t cos t 2 + C So, I = 2 x - 1 sin - 1 x 2 + 1 2 x 1 - x + C .

Integrate with respect to x : ix x 2 tan - 1 x 1 + x 2

Given I = ∫ x 2 tan - 1 x 1 + x 2 d x ⇒ I = ∫ 1 + x 2 - 1 1 + x 2 tan - 1 x d x ⇒ I = ∫ 1 · tan - 1 x d x - ∫ tan - 1 x 1 + x 2 d x Consider first function is tan - 1 x and second function is 1 and apply integration by parts in first part and assume tan - 1 x = t in second part, we get I = tan - 1 x ∫ 1 · d x - ∫ d d x tan - 1 x ∫ 1 · d x d x - ∫ t d t I = x tan - 1 x - 1 2 ∫ 2 x 1 + x 2 d x - t 2 2 ⇒ I = x tan - 1 x - 1 2 ln 1 + x 2 - tan - 1 x 2 2 + C , where C is arbitrary constant.