Mathhouse中的一些積分題

2023-07-23 22:45 作者:現(xiàn)代微積分 0人讀過 | 我要投稿

今天起興翻了翻以前的“裝b”記錄，發(fā)現(xiàn)了@Mathhouse這位up。評論可以追溯到21年9~12月那時了。嗨，想當(dāng)年我在這up評論區(qū)可謂“叱咤風(fēng)云”，我有幸也成為他首個(非官方號的)關(guān)注[滑稽]

另外，當(dāng)年混日子的事兒可就別提了，我沉迷到回宿舍刷視頻到凌晨(

我當(dāng)時做數(shù)學(xué)題起來的沉迷度可跟大多數(shù)人看小說一樣，所以....嗨，復(fù)試已經(jīng)給我深刻教訓(xùn)了，以后一定要平衡好的

可惜的是這位up不知什么原因停更一年了，期待他有朝一日繼續(xù)回來更新[滑稽]

重新做了做這位up發(fā)的題，發(fā)現(xiàn)有些題目已經(jīng)精通很多聊~(摸魚的"小成果")截取部分題來給出自己的解析

別擔(dān)心，這些都是例子，如果嫌長可以先跳到最后看總結(jié)再回頭做這些題

注：下面的積分均省略+C

(1) $%5Cint%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5Csqrt%7Bx%7D-x%5E2%20%7D%20%7D%20%5Cmathrm%7Bd%7Dx$

積分 1/Sqrt[x Sqrt[x] - x^2]

這題可以用切比雪夫定理解決

被積函數(shù)化成二項(xiàng)微分式，即 $x%5E%7B-%5Cfrac%7B3%7D%7B4%7D%20%7D(1-x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7D$

$%5Calpha%20%3D-%5Cfrac%7B3%7D%7B4%7D%2Ca%3D1%2Cb%3D-1%2C%5Cbeta%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2C%5Cgamma%20%3D-%5Cfrac%7B1%7D%7B2%7D%20$

其中 $%5Cfrac%7B%5Calpha%20%2B1%7D%7B%5Cbeta%20%7D%2B%5Cgamma%20%3D0%5Cin%20%5Cmathbb%7BZ%7D%20$ ，屬于情形三，于是令

$u%5E2%3D%5Cfrac%7B1-x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%7D%20$ ,則有： $%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0A%20x%3D%5Cfrac%7B1%7D%7B(u%5E2%2B1)%5E2%7D%20%5C%5C%0A2u%5Cmathrm%7Bd%7Du%3D-%5Cfrac%7B1%7D%7B2%7D%20x%5E%7B-%5Cfrac%7B3%7D%7B2%7D%20%7D%5Cmathrm%7Bd%7Dx%0A%5Cend%7Bmatrix%7D%5Cright.$

于是

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D-4%5Cint%20x%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7D(1-x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7Du%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D-4%5Cint%20(%5Cfrac%7B1%7D%7B(1%2Bu%5E2)%5E2%7D%20)%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7D(1-(%5Cfrac%7B1%7D%7B(1%2Bu%5E2)%5E2%7D%20)%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7Du%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D-4%5Cint%20%5Cfrac%7B1%7D%7Bu%5E2%2B1%7D%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D-4%5Ctan%5E%7B-1%7Du%5C%5C%0A%26%3D-4%5Ctan%5E%7B-1%7D%5Csqrt%7Bx%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7D-1%7D%20%0A%5Cend%7Balign%7D$

ps:第二行看似長了些，實(shí)際上只是簡單的冪指數(shù)運(yùn)算而已。

(2) $%5Cint%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx-%5Csqrt%7Bx%7D%20%7D%20%7D%20%5Cmathrm%7Bd%7Dx$

雙重根式函數(shù)積分 Integral of 1/Sqrt[x - Sqrt[x]] dx = ？

也是用切比雪夫定理

被積函數(shù)化成二項(xiàng)微分式，即 $x%5E%7B-%5Cfrac%7B1%7D%7B4%7D%20%7D(-1%2Bx%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7D$

屬于情形三，于是令 $u%5E2%3D%5Cfrac%7B-1%2Bx%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%7D%20$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D4%5Cint%20x%5E%7B%5Cfrac%7B5%7D%7B4%7D%20%7D(x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D-1)%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7Du%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D4%5Cint%20%5Cfrac%7B1%7D%7B(1-u%5E2)%5E2%7D%20%5Cmathrm%7Bd%7Du%0A%5Cend%7Balign%7D$

到此可以采用三角換元 $u%3D%5Csin%20t$ 了，這里給出另一種分部積分的做法

將分子利用1的代換化為 $1-u%5E2%2Bu%5E2$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20%5Cfrac%7B1%7D%7B1-u%5E2%7D%5Cmathrm%7Bd%7Du%2B%5Cint%20%5Cfrac%7Bu%5E2%7D%7B(1-u%5E2)%5E2%7D%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D%5Ccosh%5E%7B-1%7Du%2B%5Cint%5Cfrac%7B1%7D%7B2%7D%20u%5Cmathrm%7Bd%7D(%5Cfrac%7B1%7D%7B1-u%5E2%7D%20)%5C%5C%0A%26%3D%5Ccosh%5E%7B-1%7Du%2B%5Cfrac%7B1%7D%7B2%7D%20%20%5Cfrac%7Bu%7D%7B1-u%5E2%7D-%5Cfrac%7B1%7D%7B2%7D%5Cint%20%5Cfrac%7B1%7D%7B1-u%5E2%7D%5Cmathrm%7Bd%7Du%20%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ccosh%5E%7B-1%7Du%2B%5Cfrac%7B1%7D%7B2%7D%20%20%5Cfrac%7Bu%7D%7B1-u%5E2%7D%0A%5Cend%7Balign%7D$

再回代即可

當(dāng)然，這題也可以用雙元法

這個是在混知乎的時候?qū)W了點(diǎn)皮毛(方法源自虛調(diào)子)，要有一定基礎(chǔ)才行，新手姑且將p,q理解為分別代指那背后的兩個函數(shù)吧

$%5Csqrt%7Bx-%5Csqrt%7Bx%7D%20%7D%20%3D%5Csqrt%7B(%5Csqrt%7Bx%7D-%5Cfrac%7B1%7D%7B2%7D)%5E2-%5Cfrac%7B1%7D%7B4%7D%20%20%20%7D$

令 $p%3D%5Csqrt%7Bx%7D%20-%5Cfrac%7B1%7D%7B2%7D%20%2Cq%3D%5Csqrt%7Bx-%5Csqrt%7Bx%7D%20%7D%20$ ，有：

$p%5E2-q%5E2%3D%5Cfrac%7B1%7D%7B4%7D%20%5CRightarrow%20p%5Cmathrm%7Bd%7D%20p%3Dq%5Cmathrm%7Bd%7D%20q$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20%5Cfrac%7B1%7D%7Bq%7D%5Cmathrm%7Bd%7D%20%5B(p%2B%5Cfrac%7B1%7D%7B2%7D%20)%5E2%5D%20%5C%5C%0A%26%3D2%5Cint%20%5Cfrac%7Bp%7D%7Bq%7D%20%5Cmathrm%7Bd%7Dp%2B%5Cint%20%5Cfrac%7B1%7D%7Bq%7D%20%5Cmathrm%7Bd%7Dp%5C%5C%0A%26%3D2%5Cint%20%5Cmathrm%7Bd%7Dq%2B%5Cint%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bp%5E2-%5Cfrac%7B1%7D%7B4%7D%20%7D%20%7D%20%5Cmathrm%7Bd%7Dp%5C%5C%0A%26%3D2q%2B%5Ccosh%5E%7B-1%7D2p%5C%5C%0A%26%3D2%5Csqrt%7Bx-%5Csqrt%7Bx%7D%20%7D%20%2B%5Ccosh%5E%7B-1%7D(2%5Csqrt%7Bx%7D%20-1)%0A%5Cend%7Balign%7D$

(3) $%5Cint%20%5Cfrac%7Bx%5E2%7D%7B(x%5Csin%20x%2B%5Ccos%20x)%5E2%7D%5Cmathrm%7Bd%7Dx$

有點(diǎn)難度的積分 Integral of x^2/(x sinx + cosx)^2 dx

這里給出一個比較妙的方法：湊輔助角

對于出現(xiàn) $x%5Csin%20x%2B%5Ccos%20x%2Cx%5Csin%20x-%5Ccos%20x$ 這種玩意可以考慮用

$%5Cbegin%7Balign%7D%0A%26x%5Csin%20x%2B%5Ccos%20x%5C%5C%0A%3D%26%5Csqrt%7B1%2Bx%5E2%7D(%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%20%5Csin%20x%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%20%5Ccos%20x)%20%5C%5C%0A%3D%26%5Csqrt%7B1%2Bx%5E2%7D%5Ccos%20(x-%5Ctan%5E%7B-1%7Dx)%0A%5Cend%7Balign%7D$

于是 $%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D%5Cint%20%5Cfrac%7Bx%5E2%7D%7B(1%2Bx%5E2)%5Ccos%5E2(x-%5Ctan%5E%7B-1%7Dx)%7D%5Cmathrm%7Bd%7Dx$

注意到 $%5Cmathrm%7Bd%7D(x-%5Ctan%5E%7B-1%7Dx)%3D%5Cfrac%7Bx%5E2%7D%7Bx%5E2%2B1%7D%5Cmathrm%7Bd%7Dx$

于是

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20%5Csec%5E2(%7B%5Ccolor%7BRed%7D%20%7Bx-%5Ctan%5E%7B-1%7Dx%7D%7D%20)%5Cmathrm%7Bd%7D(%7B%5Ccolor%7BRed%7D%20%7Bx-%5Ctan%5E%7B-1%7Dx%7D%7D%20)%5C%5C%0A%26%3D%5Ctan%20(%7B%5Ccolor%7BRed%7D%20%7Bx-%5Ctan%5E%7B-1%7Dx%7D%7D)%0A%5Cend%7Balign%7D$

ps:上述法參考于無理函數(shù)君的解答

另外一個視頻是變式：求不定積分 ∫(x^2+20)/(xsin(x)+5cos(x))^2 dx

也是將分母配輔助角公式，留給讀者當(dāng)練習(xí)吧[滑稽]

(4) $%5Cint%20%5Cfrac%7B1%7D%7Bx%5E2%5Csqrt%5B4%5D%7B(x%5E4%2B1)%5E3%7D%20%7D%20%5Cmathrm%7Bd%7Dx$

有點(diǎn)難度的積分 Integral of 1/(x^2 Power[(x^4 + 1)^3, (4)^-1]) dx

這就又是用切比雪夫定理啦，跟前兩題是一個題型

被積函數(shù)化成二項(xiàng)微分式，即 $x%5E%7B-2%7D(x%5E4%2B1)%5E%7B-%5Cfrac%7B3%7D%7B4%7D%20%7D$

屬于情形三，于是令 $u%5E4%3D%5Cfrac%7B1%2Bx%5E4%7D%7Bx%5E4%7D%20$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D-%5Cint%20x%5E3(x%5E4%2B1)%5E%7B-%5Cfrac%7B3%7D%7B4%7D%20%7Du%5E3%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D-%5Cint%20%20%5Cmathrm%7Bd%7Du%5C%5C%0A%26%3D-u%5C%5C%0A%26%3D-%5Cfrac%7B%5Csqrt%5B4%5D%7B1%2Bx%5E4%7D%20%7D%7Bx%7D%20%0A%5Cend%7Balign%7D$

在視頻有點(diǎn)難度的積分Integral of (x^6 + x^3) Power[x^3 + 2, (3)^-1] dx評論區(qū)有網(wǎng)友提及了一個積分： $%5Cint%20(x%5E3-3)%5E%7B-%5Cfrac%7B1%7D%7B3%7D%20%7D%5Cmathrm%7Bd%7Dx$

就也是用切比雪夫定理，屬于情形三

令 $%5Cfrac%7Bx%5E3-3%7D%7Bx%5E3%7D%20%3Du%5E3$ ，最終化為： $%5Cint%20%5Cfrac%7Bt%7D%7B1-t%5E3%7D%5Cmathrm%7Bd%7Dt%20%20$

裂項(xiàng)后逐項(xiàng)積分再回代即可

還有幾道類似的題當(dāng)練習(xí)

求不定積分 ∫x Sqrt[1 - x^4] dx

求不定積分 Integral of x^5/Sqrt[x^3 + 1] dx

求 $%5Cint%20(x%5E2%2B1)%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%5Cmathrm%7Bd%7Dx$

評論區(qū)里的積分 Integral of (x^2 + 1)^(3/2) dx

這題有常規(guī)的三角換元 $x%3D%5Ctan%20t$ 的做法和雙曲換元 $x%3D%5Csinh%20t$ 的做法。

而如果這題用雙元法的話會爽快很多

前置結(jié)論：（圖片）

令 $y%3D%5Csqrt%7Bx%5E2%2B1%7D%20$ ，則 $y%5E2-x%5E2%3D1$

$%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D%5Cint%20y%5E3%5Cmathrm%7Bd%7Dx$

遞推式賦值 $%7B%5Ccolor%7BRed%7D%20%7Bp%7D%7D%20%5Cto%20y%2C%7B%5Ccolor%7BDodgerBlue%7D%20%7Bq%7D%7D%20%5Cto%20x%2C%5Clambda%20%3D1$

賦值 $n%3D3$ 得：

$4%5Cint%20y%5E3%5Cmathrm%7Bd%7Dx%3Dy%5E3x%2B3%5Cint%20y%5Cmathrm%7Bd%7Dx$ ①

賦值 $n%3D1$ 得：

$2%5Cint%20y%5Cmathrm%7Bd%7Dx%3Dyx%2B%5Cint%20y%5E%7B-1%7D%5Cmathrm%7Bd%7Dx$ ②

其中

$%5Cint%20y%5E%7B-1%7D%5Cmathrm%7Bd%7Dx%3D%5Cint%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2%2B1%7D%20%7D%5Cmathrm%7Bd%7Dx%3D%5Csinh%5E%7B-1%7Dx$ ③

③代入②，②代入①得：

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20y%5E3%5Cmathrm%7Bd%7Dx%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B4%7D%20xy%5E3%2B%5Cfrac%7B3%7D%7B8%7D%20xy%2B%5Cfrac%7B3%7D%7B8%7D%5Csinh%5E%7B-1%7Dx%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B4%7D%20x(1%2Bx%5E2)%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%2B%5Cfrac%7B3%7D%7B8%7D%20x%5Csqrt%7B1%2Bx%5E2%7D%20%2B%5Cfrac%7B3%7D%7B8%7D%5Csinh%5E%7B-1%7Dx%20%5C%5C%0A%5Cend%7Balign%7D$

ps:雙元點(diǎn)火公式推導(dǎo)參考筆者在知乎寫的一篇小水文：

https://zhuanlan.zhihu.com/p/641800651

$%5Cint%20%5Cfrac%7Bx%5E3%7D%7B(1-x%5E2)%5E5%7D%20%5Cmathrm%7Bd%7Dx$

網(wǎng)友分享的積分 Integral of x^3/(1 - x^2)^5 dx

令 $y%3D%5Csqrt%7B1-x%5E2%7D%20$ ，則

$x%5E2%2By%5E2%3D1%5CRightarrow%20x%5Cmathrm%7Bd%7Dx%3D-y%5Cmathrm%7Bd%7Dy$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20%5Cfrac%7Bx%5E3%7D%7By%5E%7B10%7D%7D%5Cmathrm%7Bd%7Dx%3D-%5Cint%20%5Cfrac%7Bx%5E2%7D%7By%5E%7B9%7D%7D%5Cmathrm%7Bd%7Dy%5C%5C%0A%26%3D-%5Cint%20%5Cfrac%7B1-y%5E2%7D%7By%5E%7B9%7D%7D%5Cmathrm%7Bd%7Dy%3D%5Cint%20-%5Cfrac%7B1%7D%7By%5E9%7D%20%2B%5Cfrac%7B1%7D%7By%5E7%7D%5Cmathrm%7Bd%7Dy%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B8y%5E8%7D-%5Cfrac%7B1%7D%7B6y%5E6%7D%20%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B8(1-x%5E2)%5E4%7D-%5Cfrac%7B1%7D%7B6(1-x%5E2)%5E3%7D%0A%5Cend%7Balign%7D$

ps:其他方法抑或如評論區(qū)所言，把一個x方后面湊x^2，再湊1-x^2

再如積分對比：1/(x^2 Sqrt[1 + x^2]) vs x^2/Sqrt[1 + x^2]

簡單積分：Integral of Sqrt[x]/(1 + Power[x, (3)^-1]) dx

簡單積分：Integral of 1/(Sqrt[x] + Power[x, (3)^-1]) dx

這些都可以用切比雪夫定理求解，也可以嘗試用雙元法

到此阿B的限制的公式數(shù)居然還沒滿，那就再水幾道題...

$%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B1%7D%7B(x%2B%5Csqrt%7B1%2Bx%5E2%7D%20)%5E2%7D%5Cmathrm%7Bd%7Dx$

積分 1/(x + Sqrt[1 + x^2])^2，從 0 到 ∞

這題用雙曲換元應(yīng)該是最快的

令 $x%3D%5Csinh%20t$

則

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%20%5Cfrac%7B1%7D%7B(%5Csinh%20t%2B%5Ccosh%20t)%5E2%7D%20%5Ccdot%20%5Ccosh%20t%5Cmathrm%7Bd%7Dt%20%5C%5C%0A%26%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%20%5Cfrac%7B1%7D%7Be%5E%7B2t%7D%7D%20%5Ccdot%20%5Cfrac%7Be%5Et%2Be%5E%7B-t%7D%7D%7B2%7D%20%5Cmathrm%7Bd%7Dt%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%20%7D%20e%5E%7B-t%7D%2Be%5E%7B-3t%7D%20%5Cmathrm%7Bd%7Dt%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B2%7D(-e%5E%7B-t%7D-%5Cfrac%7B1%7D%7B3%7De%5E%7B-3t%7D%20)%7C%20_%7B0%7D%5E%7B%5Cinfty%20%7D%5C%5C%0A%26%3D%5Cfrac%7B2%7D%7B3%7D%0A%5Cend%7Balign%7D$

ps：原專欄的方法為歐拉代換，也屬于脫去二次根式的一種暴力但比較普適的方法

$%5Cint%20%5Cln%20(%5Csqrt%7Bx%2B1%7D%2B%5Csqrt%7Bx%7D%20%20)%5Cmathrm%7Bd%7Dx$

有點(diǎn)難度的積分 Integral of ln (Sqrt[x + 1] + Sqrt[x]) dx

雙曲換元，令 $x%3D%5Csinh%5E2t$

$%5Cbegin%7Balign%7D%0A%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cint%20%5Cln%20(%5Ccosh%20t%2B%5Csinh%20t)%5Cmathrm%7Bd%7D(%5Csinh%5E2t)%5C%5C%0A%26%3Dt%5Cmathrm%7Bd%7D(%5Csinh%5E2t)%3Dt%5Csinh%5E2t-%5Cint%20%5Csinh%5E2t%5Cmathrm%7Bd%7Dt%20%5C%5C%0A%26%3Dt%5Csinh%5E2t-%20%5Cint%20%5Cfrac%7B%5Ccosh2t-1%7D%7B2%7D%20%5Cmathrm%7Bd%7Dt%20%5C%5C%0A%26%3Dt%5Csinh%5E2t-%20%5Cfrac%7B1%7D%7B4%7D%5Csinh%202t%2B%5Cfrac%7B1%7D%7B2%7Dt%5C%5C%0A%26%3Dx%5Csinh%5E%7B-1%7D%5Csqrt%7Bx%7D-%20%5Cfrac%7B1%7D%7B2%7D%5Csinh%20t%5Ccosh%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Csinh%5E%7B-1%7D%5Csqrt%7Bx%7D%5C%5C%0A%26%3D(x%2B%5Cfrac%7B1%7D%7B2%7D%20)%5Csinh%5E%7B-1%7D%5Csqrt%7Bx%7D-%20%5Cfrac%7B1%7D%7B2%7D%5Csqrt%7Bx%7D%5Csqrt%7B1%2Bx%7D%20%0A%5Cend%7Balign%7D$