第四章总复习题课

选择题

1.设I=∫\frac{dx}{(e^x+e^{-x})},则I=(\quad B\quad).

A.e^x-e^{-x}+C

B.arctan\ e^x+C

C.arctan\ e^{-x}+C

D.e^x+e^{-x}+C

答案:B.arctan\ e^x+C

解析:

u=e^x;\int \frac{1}{u^2+1}du=arctan \ u;\\

\begin{aligned} I &= \int \frac{dx}{e^x+e^{-x}}\\ &=\int\frac{e^{x}}{(e^x)^2+1}dx\\ &=\int \frac{d(e^x)}{(e^{x})^2+1}\\ & =\large{\arctan\ e^x+C} \end{aligned}

2. \Large{\int \frac1{\sqrt x cos^2\sqrt x}}dx=(\ B\ )

A.\frac{1}{2} \ tan\sqrt x+C

B.2\ tan\sqrt x+C

C.-\frac12\ tan\sqrt x+C

D.-2\ tan\sqrt x+C

答案:B.2\ tan\sqrt x+C

解析:\begin{aligned} \int\frac 1{\sqrt xcos^2\sqrt x}&= 2\int \frac1{2\sqrt x \cdot cos^2\sqrt x}dx \\ &= 2\int\frac1{cos^2\sqrt x}d(\sqrt x)\\ &=2\ tan\sqrt x+C \end{aligned}

3.\int x \cdot e^{-x}dx=(\ A \ )

A.-(x+1)e^{-x}+C

B.(x+1)e^{-x}+C

C.(x-1)e^{-x}+C

D.-(x-1)e^{-x}+C

答案:A.-(x+1)e^{-x}+C

解析:

u=x; v'=e^{-x}

u'=1;v=-e^{-x}\\

\int udv=uv-\int vdu \iff \int uv'dx=uv-\int u'vdx

\begin{aligned} \int x\cdot e^{-x}dx &=\int x\cdot d(-e^{-x})\\ &=-xe^{-x}-\int-e^{-x}dx\\ &=-xe^{-x}+\int e^{-x}dx\\ &=-xe^{-x}-e^{-x}+C\\ &=-(x+1)e^{-x}+C \end{aligned}

4.\int x sinx dx=(\ A\ )

A.sinx-xcosx+C

B.sinx+xcosx+C

C.-sinx+xcosx+C

D.-sinx-xcosx+C

答案:A.sinx-xcosx+C

解析:

u=x;dv=d(-cosx)\\ du=dx;v=-cosx

\begin{aligned} \int xsinxdx&=\int xd(-cosx)\\ &=-x cosx-\int-cosxdx\\ &=-xcosx+\int cosxdx\\ &=-xcosx+sinx+C \end{aligned}

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