Tìm T = nguyên hàm x^n / 1+x+x^2/2! + x^3/ 3! +....+ x^n/ n! dx?

Câu hỏi:

A. $T=x.n!+n!\ln \left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}\right)+C$

B. $T=x.n!-n!\ln \left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}\right)+C$

C. $T=n!\ln \left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}\right)+C$

D. $T=n!\ln \left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}\right)-x^n.n!+C$

Trả lời:

Hướng dẫn:

Đặt $g\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{...}+\frac{x^n}{n!}\Rightarrow g^{\text{'}}\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathrm{...}+\frac{x^{n-1}}{\left(n-1\right)!}$

Ta có : $g\left(x\right)-g^{\text{'}}\left(x\right)=\frac{x^n}{n!}\Rightarrow x^n=n!\left(g\left(x\right)-g^{\text{'}}\left(x\right)\right)$

$\Rightarrow T=\int \frac{n!.\left[g\left(x\right)-g^{\text{'}}\left(\right)\right]}{g\left(x\right)}dx=n!\int \left[1-\frac{g^{\text{'}}\left(x\right)}{g\left(x\right)}\right]dx=n!.x-n!\ln =n!x-n!\ln \left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}\right)+C$

Vậy đáp án đúng là đáp án B .