Question

求不定积分！！！详细步骤！谢谢

∫x^2e^x^3+1dx
x的2次幂乘以e的x的3次幂加1的不定积分！~详细步骤~谢谢啦~
那个1是跟在x的三次方后面 一起作为e的指数的，不是独立的。。。。。
答案中没有x

Answer 1

解：
1.
∫ 1 / (1 + e^x) d x
= ∫ 1 / (1 + t) d ln x
= ∫ 1 / [t (1 + t)] d x
= ∫ [1 / t - 1 / (1 + t)] d x
= ∫ 1 / t d x - ∫ 1 / (1 + t) d x
= ln t - ln (t + 1) + C
= ln [t / (t + 1)] + C
= ln [e^x / (e^x + 1)] + C
2.
∫ cot x / √sin x d x
= ∫ cos x / (sin x √sin x) d x
= ∫ 1 / (sin x √sin x) d sin x
= ∫ 1 / (t √t) d t
= - 2 ∫ t^(- 3 / 2) d t
= - 2 t^(- 1 / 2) + C
= - 2 (sin x)^(- 1 / 2) + C
3.
∫ (sec x)^2 / √(1 + tan x) d x
= ∫ 1 / √(1 + tan x) d tan x
= ∫ 1 / √(1 + t) d t
= 2 √(1 + t) + C
= 2 √(1 + tan x) + C
4.
∫ (arctan x)^2 / (1 + x^2) d x
= ∫ t^2 / [1 + (tan t)^2] d tan t
= ∫ t^2 / {(cos t)^2 [1 + (tan t)^2]} d t
= ∫ t^2 / [(cos t)^2 + (sin t)^2] d t
= ∫ t^2 d t
= t^3 / 3 + C
= (arctan x)^3 / 3 + C
5.
∫ x (x^2 - 1)^(1 / 3) d x
= ∫ 1 / 2 (x^2 - 1)^(1 / 3) d x^2
= 1 / 2 ∫ (t - 1)^(1 / 3) d t
= 1 / 2 * 3 / 4 (t - 1)^(4 / 3) + C
= 3 / 8 (x^2 - 1)^(4 / 3) + C
6.
∫ x ln x d x
= ∫ e^t t d e^t
= ∫ e^(2 t) t d t
= 1 / 4 ∫ e^(2 t) 2 t d (2 t)
= 1 / 4 ∫ e^u u d u
= 1 / 4 ∫ u d e^u
= 1 / 4 (u e^u - ∫ e^u du + C1)
= 1 / 4 (u e^u - e^u + C2)
= (u - 1) e^u / 4 + C
= (2 ln x - 1) e^(ln x^2) / 4 + C
= x^2 (2 ln x - 1) / 4 + C
7.
∫ x^2 e^x d x
= ∫ x^2 d e^x
= x^2 e^x - ∫ e^x d x^2 + C1
= x^2 e^x - ∫ 2 x d e^x + C1
= x^2 e^x - 2 x e^x + ∫ e^x d 2 x + C2
= x^2 e^x - 2 x e^x + 2 e^x + C
8.
∫ arcsin x d x
= ∫ t d sin t
= t sin t - ∫ sin t d t + C1
= t sin t + cos t + C
= t sin t + √[1 - (sin t)^2] + C
= x arcsin x + √(1 - x^2) + C
9.
∫ (x + 1) / [x √(x - 2)] d x
= ∫ (2 + t^2 + 1) / [(2 + t^2) t] d (2 + t^2)
= 2 ∫ t (2 + t^2 + 1) / [(2 + t^2) t] d t
= 2 ∫ (2 + t^2 + 1) / (2 + t^2) d t
= 2 ∫ [1 + 1 / (2 + t^2)] d t
= 2 { t + C1 + ∫ 1 / [(√2)^2 + t^2] d t }
= 2 [t + C1 + 1 / √2 arctan (t / √2) + C2]
= 2 t + √2 arctan (√2 t / 2) + C
= 2 √(x - 2) + √2 arctan [√(2 x - 4) / 2] + C
10.
∫ √(x^2 - 9) / x d x
= 1 / 2 ∫ √(x^2 - 9) / x^2 d x^2
= 1 / 2 ∫ √(t - 9) / t d t
= 1 / 2 ∫ u / (u^2 + 9) d (u^2 + 9)
= ∫ u^2 / (u^2 + 9) d u
= ∫ 1 d u - 9 ∫ 1 / (u^2 + 9) d u
= u + C1 - 9 * 1 / 3 * arctan (u / 3) + C2
= u + 3 arctan (u / 3) + C
= √(x^2 - 9) + 3 arctan (√(x^2 - 9) / 3) + C
11.
∫ (x^3 + 1) / (x^3 + x) d x
= ∫ { 1 + (1 - x) / [x (x^2 + 1)] } d x
= ∫ [ 1 + 1 / x - (x + 1) / (x^2 + 1) ] d x
= x + ln x + C1 - ∫ x / (x^2 + 1) d x - ∫ 1 / (x^2 + 1) d x
= x + ln x + C1 - 1 / 2 ∫ 1 / (x^2 + 1) d x^2 - ∫ 1 / (x^2 + 1) d x
= x + ln x + C1 - 1 / 2 ∫ 1 / (t + 1) d t - ∫ 1 / (x^2 + 1) d x
= x + ln x + C1 - [ln (t + 1)] / 2 + C2 - ∫ 1 / (x^2 + 1) d x
= x + ln x + C1 - [ln (x^2 + 1)] / 2 + C2 - arctan x + C3
= x + ln x - [ln (x^2 + 1)] / 2 - arctan x + C
12.
∫ 5^x e^x d x
= ∫ (5 e)^x d x
= ∫ 1 / ln (5 e) d (5 e)^x
= 1 / ln (5 e) * (5 e)^x
= (5 e)^x / (1 + ln 5)

Answer 2

∫x^2e^x^3+1dx
=∫x^2e^x^3dx+∫dx
=1/3*∫e^x^3dx^3+x
=(1/3)(e^x^3)+x+C

Answer 3

∫x^2e^x^3+1dx
=1/3∫e^x^3+1dx^3
=1/3(e^x^3+1)^2+C