Chứng minh rằng: $\int\limits {\sqrt{x^2+h}} \, dx=\frac{x}{2}\sqrt{x^2+h}+\frac{h}{2}ln|x+\sqrt{x^2+h}|+C$
2 câu trả lời
Lời giải:
Đặt $\left \{ {{u=\sqrt{x^2+h}} \atop {dv=dx}} \right.=>$$\left \{ {{du=\frac{x}{\sqrt{x^2+h}}dx} \atop {v=x}} \right.$
=>$I=x\sqrt{x^2+h}-$$\int\limits {\frac{x^2}{\sqrt{x^2+h}}} \, dx$
Ta có:
$I_1=$$\int\limits {\frac{x^2}{\sqrt{x^2+h}}} \, dx=$ $\int\limits {\frac{x^2+h-h}{\sqrt{x^2+h}}} \, dx=$ $\int\limits {\sqrt{x^2+h}} \, dx-h$ $\int\limits {\frac{dx}{\sqrt{x^2+h}}} \, =I-hln|x+\sqrt{x^2+h}+C_1|$
=> $\int\limits {\sqrt{x^2+h}} \, dx=\frac{x}{2}\sqrt{x^2+h}+\frac{h}{2}ln|x+\sqrt{x^2+h}|+C$
Đáp án:
Đặt $\left \{ {{u=\sqrt{x^2+h}} \atop {dv=dx}} \right.=>$$\left \{ {{du=\frac{x}{\sqrt{x^2+h}}dx} \atop {v=x}} \right.$
=>$I=x\sqrt{x^2+h}-$$\int\limits {\frac{x^2}{\sqrt{x^2+h}}} \, dx$
Ta có:
$I_1=$$\int\limits {\frac{x^2}{\sqrt{x^2+h}}} \, dx=$ $\int\limits {\frac{x^2+h-h}{\sqrt{x^2+h}}} \, dx=$ $\int\limits {\sqrt{x^2+h}} \, dx-h$ $\int\limits {\frac{dx}{\sqrt{x^2+h}}} \, =I-hln|x+\sqrt{x^2+h}+C_1|$
=> $\int\limits {\sqrt{x^2+h}} \, dx=\frac{x}{2}\sqrt{x^2+h}+\frac{h}{2}ln|x+\sqrt{x^2+h}|+C$
Chúc bạn học tốt!!!
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