[imath]\displaystyle \int^{\ln{6}}_{0}{ \dfrac{ e ^{x}}{1+ \sqrt{e^x+3}}} \mathrm{d} x[/imath]
Đặt [imath]t= \sqrt{e^x+3} \ (t>0) \Rightarrow t^2=e^x+3 \\
\Rightarrow 2t\mathrm{d} t= e^x \mathrm{d} x[/imath]
Đổi cận:
[imath]x=0 \Rightarrow t=2 \\
x= \ln{6} \Rightarrow t=3[/imath]
Ta có: [imath]\displaystyle \int^{3}_{2}{ \dfrac{ 2t}{1+ t}} \mathrm{d} t = 2 \displaystyle \int^{3}_{2}{ \dfrac{ t}{1+ t}} \mathrm{d} t \\
\ \ \ =2 \displaystyle \int^{3}_{2}{ \left ( 1 - \dfrac{ 1}{1+ t} \right ) }\mathrm{d} t \\
=\displaystyle 2 \left( \int^{3}_{2} \mathrm{d} t - \int^{3}_{2}{ \dfrac{ \mathrm{d} t}{1+ t} } \right ) \\
= 2 [(t)|^3_2 - (\ln|t+1|)|^3_2] \\
=2.[1-( \ln{4} - \ln{3})] \\
=2 - 2 \ln{4} + 2 \ln{3} \\
= 2 - 2 \ln{(2^2)} + 2 \ln{3} \\
= 2 - 4 \ln{2} + 2 \ln{3}[/imath]
[imath]\displaystyle \int^{\ln{6}}_{0}{ \dfrac{ e ^{x}}{1+ \sqrt{e^x+3}}} \mathrm{d} x = a+ b \ln{2} + c \ln{3}[/imath]
Suy ra [imath]a=2, \ b = -4, \ c = 2[/imath]
[imath]\Rightarrow T=a+b+c=0[/imath]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
