Tìm nguyên hàm e^x( 3x-2)+ căn x-1/ căn x-1 ( e^x. căn x-1+1)dx ?

Câu hỏi:

Tìm

I = \int \frac{e^{x} (3 x - 2) + \sqrt{x - 1}}{\sqrt{x - 1} (e^{x} . \sqrt{x - 1} + 1)} d x

A. $I = x + \ln (e^{x} . \sqrt{x - 1} + 1) + C$

B. $I = x - \ln (e^{x} . \sqrt{x - 1} + 1) + C$

C. $I = \ln (e^{x} . \sqrt{x - 1} + 1) + C$

D. $I = \ln (e^{x} . \sqrt{x - 1} - 1) + C$

Trả lời:

Hướng dẫn:

$I = \int \frac{e^{x} (3 x - 2) + \sqrt{x - 1}}{\sqrt{x - 1} (e^{x} . \sqrt{x - 1} + 1)} d x = \int \frac{\sqrt{x - 1} (e^{x} . \sqrt{x - 1} + 1) + e^{x} (2 x - 1)}{\sqrt{x - 1} (e^{x} . \sqrt{x - 1} + 1)} d x = \int d x + \int \frac{e^{x} (2 x - 1)}{\sqrt{x - 1} (e^{x} . \sqrt{x - 1} + 1)} d x$

Đặt : $t = e^{x} . \sqrt{x - 1} + 1 \Rightarrow d t = (\frac{e^{x}}{2 \sqrt{x - 1}} + e^{x} \sqrt{x - 1}) d x = \frac{e^{x} (2 x - 1)}{2 \sqrt{x - 1}} d x$

Vậy $\Rightarrow I = \int d x + \int \frac{e^{x} (2 x - 1)}{\sqrt{x - 1} (e^{x} \sqrt{x - 1} + 1)} d x = x + \int \frac{1}{t} d t = x + \ln |t| + C = x + \ln (e^{x} . \sqrt{x - 1} + 1) + C$