解:$$ \int \sqrt{1+e^{x}}dx $$ 令$$ t= \sqrt{1+e^{x}} $$ ∴$$ x= \tan(t^{2}-1) $$ $$ dx= \frac{2t}{t^{2}-1} $$ ∴$$ \int \sqrt{1+e^{x}}dx $$ $$ = \int \frac{2t^{2}}{t^{2}-1}dt \\ = \int \frac{2t^{2}-2+2}{t^{2}-1}d...
【解析】、$$ \int \sqrt { e ^ { x } } $$dx,令$$ t = e ^ { x } $$,则$$ \int V e ^ { x } d x = \int V t d \ln t = \int $$ 1/√t$$ d t = 2 \sqrt { t } $$ 带回$$ t = e ^ { x } $$ 则$$ \int V e ^ { x } d x = 2 \sqrt {...
The tone of the question draw me to think that you are looking for something like "fractionnal calculus", i.e. ... Using n=4 trapezoids, how do you approximate the value of ∫x+1dx from [1,3]? https://socratic.org/questions/using-n-4-trapezoids-how-do-you-approximate-the-value...
Explanation: Let I=∫e2x−91dx We ... How to integrate ∫1+e2xdx https://www.quora.com/How-do-you-integrate-int-frac-dx-sqrt-1+e-2x Substitute y = e^x, so dx = dy/y, and then substitute y = sh z (sin hyperbolicus), to get rid of sqrt(1+y^...
首先设e^x=t 则x=lntdx=1/tdt1/sqrt(1+e^x) dx=1/[t*sqrt(1+t)]dt在设sqrt(1+t)=m t=m^2-1dt=2mdm1/[t*sqrt(1+t)]dt=2m/[(m^2-1)*m]dm=2/[(m+1)(m-1)]dm=[1/(m-1)-1/(m+1)]dm=ln(m-1)-ln(m+1)=ln[sqrt(1+e^x)-1]-l... 解析看不懂?免费查看同类...
$$ 令$$ \sqrt { e ^ { x } - 1 } = $$,则$$ x = \ln ( t ^ { 2 } + 1 ) d x = \frac { 2 t } { t ^ { 2 } + 1 } $$dr,于是 $$ \int \sqrt { e ^ { x } - 1 } d x = \int t \cdot \frac { 2 t } { t ^ { 2 } + 1 } d t ...
u + 1 | - \frac { 1 } { 2 ( u + 1 ) } + c \\ = \frac { 1 } { 2 } \ln | \frac { \sqrt { 1 + e ^ { x } } - 1 } { \sqrt { 1 + e ^ { x } } + 1 } | - \frac { \sqrt { 1 + e ^ { x } } } { 2 e ^ { x } } + c _ {...
当然有了，如果说广义傅里叶变换指的是其作为广义缓增函数的傅里叶变换，那么由于它局部可积并且在无穷远处以多项式速度增长所以一定存在傅里叶变换，只是它的傅里叶变换可能不是一般的函数而是广义函数。F[x]=limR→∞∫−RRxe−2πix⋅ξdx，但是当x<0时x=i−x 为了方便首先对ξ求积分，可以验证...
用对数求导法求函数y=\sqrt{x\sin x\sqrt{1-e^x}}的导数. 解: 因为 \begin{aligned} \ln y=&\frac{1}{2}\ln(x\sin x\sqrt{1-e^x})\\ =&\frac{1}{2}[\ln x+\ln\sin x+\frac{1}{2}\ln(1-e^x)]\\ =&\frac{1}{2}\ln x+\frac{1}{2}\ln\sin x+\frac{1}{4}\...
How do you integrate ∫x2x+1 using substitution? https://socratic.org/questions/how-do-you-integrate-int-xsqrt-2x-1-using-substitution =101(2x+1)25−61(2x+1)23+C Explanation: Let u=2x+1⇒du=2dx ... How do you integrate ∫5x2x+3 using substitution? https://socratic.org/question...