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\begin{document}
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1. \[\int e^{\sin^2 x}\sin 2x \,dx=\int
e^{\sin^2x}d\sin^2x=e^{\sin^2x}\]

проверка

\[(e^{\sin^2x})'=e^{\sin^2x}\,2\sin x\cos x=e^{sin^2x}\sin 2x\]

2. \[\int \arctg\sqrt{x}\,dx\]

берем по частям, $u=\arctg\sqrt{x},\quad v=x$
\[I=x\, \arctg\sqrt{x}-\int x\frac
1{x+1}\frac 1{2\sqrt{x}}\,dx=x \,\arctg\sqrt{x}-\frac
12\int\frac{\sqrt{x}}{x+1}\,dx\]

делаем замену $x=z^2\quad\to\quad$  второй интеграл примет вид

\[\int\frac{z^2}{1+z^2}\,dz=\int\frac{z^2+1-1}{z^2+1}\,dz=
z-\arctg\,z=\sqrt{x}-\arctg\sqrt{x}\]

Ответ \[I=(x+1)\arctg\sqrt{x}-\sqrt{x}\]

Проверка \[\arctg\sqrt{x}+\frac{1+x}{1+x}\frac 1{2\sqrt{x}}-\frac
1{2\sqrt{x}}\]

3. \[\int\frac{dx}{x^3+8}=\int\frac{dx}{(x+2)(x^2-2x+4)}\]

разлагаем дробь на две части

\[\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4},\quad
A(x^2-2x+4)+(Bx+C)(x+2)=1\quad\to\quad\]\[
\left\{\begin{array}{l}
A+B=0\\ -2A+2B+C=0\\ 4+2C=1\end{array}\right.
\quad\to\quad
A=\frac 1{12},\quad B=-\frac 1{12},\quad C=\frac 4{12}\]

\[I=\frac 1{12}\int\left(\frac 1{x+2}+\frac{-x+4}{x^2-2x+4}\right)\,dx\]

\[I_1=\log|x+2|,\qquad I_2=\int\frac{x-4}{(x-1)^2+3}\,dx\]

замена $y=x-1$
\[I_2=\int\frac{y-3}{y^2+3}\,dy=
\int\frac{y\,dy}{y^2+3}-3\int\frac{dy}{y^2+3}=
\frac 12\int\frac{dy^2}{y^2+3}-\int\frac{dy}{\frac{y^2}3+1}=
\frac 12\log|y^2+3|-\int\frac{dy}{\left(\frac
y{\sqrt{x}}\right)^2+1}=\]\[
...-\sqrt{3}\int\frac{d\frac{y}{\sqrt{3}}}
{\left(\frac{y}{\sqrt{3}}\right)^2+1}=
...-\sqrt{3}\,\arctg\frac
y{\sqrt{3}}=\log\sqrt{x^2-2x+4}-\sqrt{3}\,\arctg\frac{x-1}{\sqrt{3}}\]

Ответ \[\frac 1{12}\left(\log|x+2|-\log\sqrt{x^2-2x+4}+
\sqrt{3}\arctg\frac{x-1}{\sqrt{3}}\right)\]

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4. \[\int\frac{dx}{1+\sqrt[3]{x+1}}\] замена
$z=\sqrt[3]{x+1}$\[
I=\int\frac{3z^2\,dz}{1+z}=3\int\frac{z^1-1+1}{z+1}\,dz=
3\left(\int\frac{z^2-1}{z+1}\,dz+\int\frac{dz}{z+1}\right)=3\left(\int(z-1)dz
+\log|z+1|\right)=\]\[
3(z^2/2-z+\log|z+1|)=3((x+1)^{2/3}/2-\sqrt[3]{x+1}+\log|\sqrt[3]{x+1}+1|)\]

5. \[\int e^x\log(1+3e^x)\,dx\]

замена $e^x=y$

\[I=\int y\log(1+3y)\frac 1y\,dy=\int\log(1+3y)\,dy\]

замена $z=1+3y$

\[I=\frac 13\int\log z\,dz=\frac 13(z\log z-z)=\frac 13z\log z-1)=\frac
13(1+3e^x)(\log(1+3e^x)-1)\]

6.
\[\int\frac{2x^2-3x+1}{x^3+1}\,dx=
\int\frac{2x^2-2x+2-(x+1)}{(x+1)(x^2-x+1)}\,dx=
2\int\frac{dx}{x+1}-\int{dx}{x^2-x+1}=\]\[2\log|x+1|-
\int\frac{dx}{\left(x-\frac 12\right)^2+\frac 34}\]

замена $y=x-\frac 12$\[ I_2=\int\frac{dy}{y^2+\frac 34}=\frac
34\int{dy}{\left(\frac{2y}{\sqrt{3}}\right)^2+1}=
\frac
43\frac{\sqrt{3}}2\int\frac{d\frac{2y}{\sqrt{3}}}{\left(\frac{2y}
{\sqrt{3}}\right)^2+1}=
\frac 2{\sqrt{3}}\,\arctg\frac{2y}{\sqrt{3}}=\frac
2{\sqrt{3}}\,\arctg\frac{2x-1}{\sqrt{3}}\]

Ответ
\[I=2\log|x+1|-\frac 2{\sqrt{3}}\arctg\frac{2x-1}{\sqrt{3}}\]

7. \[\int\frac{x\,dx}{(x^2+4)^6}=\frac 12\int\frac{dx^2}{(x^2+4)^6}=
\frac 12\int(x^2+4)^{-6}d(x^2+4)=\frac 12(x^2+4)^{-5}/(-5)=-\frac
1{10(x^2+4)^5}\]

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8.
\[ \int x\,3^x\,dx\]

берем по частям, $u=x,\quad v=3^x/\log 3$

\[ I=\frac{x\,3^x}{\log 3}-\int\frac{3^x}{\log 3}\,dx=\frac{3^x}{\log
3}\left(x-\frac 1{\log 3}\right)\]

9.
\[ \int x^2\,e^{3x}\,dx\]

берем по частям,  $u=x^2,\quad v=e^{3x}/3$

\[ I=x^2\,e^{3x}/3-\frac 23\int x\,e^{3x}\,dx\]

берем по частям, $u=x,\quad v=e^{3x}/3$

\[ I=x^2\,e^{3x}/3-\frac 23\left(x\,e^{3x}/3-\frac
13\int e^{3x}\,dx\right)=
... \left( ...-\frac 13 \frac{e^{3x}}3\right)=
\frac {e^{3x}}{3}\left(x^2-\frac 23x+\frac 29\right)\]

10.
\[ \int\frac{\cos 3x\,dx}{4+\sin 3x}=\frac 13\int\frac{\cos 3x\,d3x}
{4+\sin 3x}\]

замена $y=3x$

\[ I=\frac 13\int\frac{\cos y\,dy}{\sin y+4}=\frac 13\int\frac{d\sin
y}{\sin y+4}=\frac 13\log|\sin y+4|=\frac 13\log|\sin 3x+4|\]

11.
\[ \int\frac{x\,\arcsin x\,dx}{\sqrt{1-x^2}}\]

\[ \frac{dx}{\sqrt{1-x^2}}=d\,\arcsin x,\qquad \mbox{замена}\quad
y=\arcsin x,\quad x=\sin y,\qquad I=\int y\sin y\,dy\]

берем по частям $u=y, \quad v=-\cos y$

\[ I=-y\,\cos y+\int\cos y\,dy=-y\,\cos y+\sin y=
-y\sqrt{1-\sin^2y}+\sin y=-\arcsin x\sqrt{1-x^2}+x\]

12.
\[ \int\frac{dx}{\cos^2x(3\tan x+1)}\]

\[ \frac{dx}{\cos^2x}=d\,\tan x,\qquad\mbox{замена}\qquad \tan
x=y,\]
\[ I=\int\frac{dy}{3y+1}=\frac 13\int{d(3y+1)}{3y+1}=
\frac 13\log|3y+1|=\frac 13\log|3\tan x+1|\]

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13.
\[\int\frac{3x-7}{x^3+4x^2+4x+16}\,dx=\int\frac{3x-7}{(x+4)(x^2+4)}\,dx\]

разлагаем дробь на части

\[ \frac{Ax+B}{x^2+4}+\frac{C}{x+4}=\frac{3x-7}{...}\quad\to\quad
(Ax+B)(x+4)+C(x^2+4)=3x-7\]
\[ \left\{\begin{array}{l}
A+C=0\\ 4A+B=3\\ $B+$C=-7\end{array}\right.\quad\to\quad
A=\frac{19}{20},\quad B=-\frac{16}{20},\quad C=-\frac{19}{20}\]

\[ I=\frac 1{20}\left(19\int\frac{x\,dx}{x^2+4}-16\int\frac{dx}{x^2+4}-
19\int\frac{dx}{x+4}\right)=
\frac 1{20}\left(\frac{19}2\int\frac{dx^2}
{x^2+4}-16\int\frac{dx}{\left(\left(\frac x2\right)^2+1\right)4}-
19\log|x+4|\right)=\]
\[ \frac 1{20}\left(\frac{19}2\log(x^2+4)-8\arctg \frac
x2-19\log|x+4|\right)=
\frac 1{20}\left(19\frac{\sqrt{x^2+4}}{|x+4|}-8\arctg\frac x2\right)\]

14.
\[ \int\frac{dx}{x^3+x^2+2x+2}=\int\frac{dx}{(x^2+2)(x+1)}\]

разлагаем дробь

\[ \frac{Ax+B}{x^2+2}+\frac{C}{x+1}=\frac 1{...}\quad\to\quad
(Ax+B)(x+1)+C(x^2+2)=1\]
\[ \left\{\begin{array}{l}
A+C=0\\ A+B=0\\ B+2C=1\end{array}\right.\quad\to\quad
A=-\frac 13,\quad B=\frac 13,\quad C=\frac 13\]
\[ I=\frac 13\left(-\int\frac{x\,dx}{x^2+2}+\int\frac{dx}{x^2+2}+
\int\frac{dx}{x+1}\right)=
\frac 13\left(-\frac 12\log(x^2+2)+\frac 1{\sqrt{2}}\arctg\frac
x{\sqrt{2}}+\log|x+1|\right)=\]\[
\frac 13\left(\log\frac{|x+1|}{\sqrt{x^2+2}}+\frac
1{\sqrt{2}}\arctg\frac x{\sqrt{2}}\right)\]

15.
\[ \int\frac{x^2\,dx}{x^3+5x^2+8x+4}=\int\frac{dx}{(x+1)(x+2)^2}=\]

разлагаем дроби как обычно

\[ I=-4\int\frac{dx}{(x+2)^2}+\int\frac{dx}{x+1}=\frac
4{x+2}+\log|x+1|\]

16.
\[ \int\frac{\sin x\,dx}{\sqrt[3]{\cos^2x}}=-\int\frac{d\,\cos
x}{\sqrt[3]{\cos^2x}}=-\int(\cos x)^{-2/3}\,d\,\cos x=
-3(\cos x)^{1/3}=-3\sqrt[3]{\cos x}\]

17.
\[ \int\frac{(x+\arctg x)\,dx}{1+x^2}=\int\frac{x\,dx}{x^2+1}+
\int{\arctg x\,dx}{x^2+1}=
\frac 12\int\frac{d\,x^2}{x^2+1}+\int\arctg x\,d\arctg x=
\frac 12\log(x^2+1)+\frac 12\arctg^2x\]

18.
\[ \int\frac{\arctg \sqrt{x}}{\sqrt{x}(1+x)}\]

замена $\sqrt{x}=y$

\[ I=\int\frac{\arctg y\,2y\,dy}{y(1+y^2)}=2\int\frac{\arctg
y\,dy}{1+y^2}=2\int\arctg y\,d\arctg y=\arctg^2y=\arctg^2\sqrt{x}\]


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19.
\[ \int\frac{x^2\,dx}{x^4-81}=\int\frac{dx}{(x-3)(x+3)(x^2+9)}\]

разлагаем дробь на три части

\[
\frac{A}{x-3}+\frac{B}{x+3}+\frac{Cx+D}{x^2+9}=\frac{x^2}{...}\quad\to\quad
\left\{\begin{array}{l}
A+B+C=0\\ 3A-3B+D=1\\ 9A+9B-9C=0\\ 27A-27B-9D=0\end{array}\right.
\quad\to\quad \begin{array}{l}
A=1/12\\ B=-1/12\\ D=6/12\\ C=0\end{array}\]

\[ I=\frac
1{12}\left(\int\frac{dx}{x-3}-\int\frac{dx}{x+3}+6\int\frac{dx}{x^2+9}\right)=
\frac 1{12}\left(\log|x-3|-\log|x+3|+2\arctg\frac x3\right)=\]\[
\frac 1{12}\left(\log\left|\frac{x-3}{x+3}\right|+2\arctg\frac
x3\right)\]

20.
\[ \int\frac{(x+3)\,dx}{x^3+x^2-2x}=\int\frac{x+3}{x(x-1)(x+2)}\,dx\]

разлагаем дробь на три части

\[ \frac Ax+\frac B{x-1}+\frac C{x+2}=\frac{x+3}{...}\]
\[ A=-3/2,\quad B=11/2,\quad C=-4\]
\[ I=\frac
12\left(-3\int\frac{dx}x+11\int\frac{dx}{x-1}-8\int\frac{dx}{x+2}\right)=
\frac 12\left(-3\log|x|+11\log|x-1|-8\log|x+2|\right)\]

21.
\[
\int\frac{(x^2-3)\,dx}{x^4+5x^2+6}=\int\frac{x^2-3}{(x^2+3)(x^2+2)}\,dx\]
\[ \frac A{x^2+3}+\frac B{x^3+2}=\frac{x^2-3}{...}\quad\to\quad
A=6,\quad B=-5\]
\[ I=6\int\frac{dx}{x^2+3}-5\int\frac{dx}{x^2+2}=
2\sqrt{3}\int\frac{d\,\frac x{\sqrt{3}}}{\left(\frac
x{\sqrt{3}}\right)^2+1}-\frac 5{\sqrt{2}}\int\frac{d\,\frac
x{\sqrt{2}}}{\left(\frac x{\sqrt{2}}\right)^2+1}=
 2\sqrt{3}\arctg\frac x{\sqrt{3}}-\frac 5{\sqrt{2}}\arctg\frac
x{\sqrt{2}}\]


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22.
\[ \int\frac{\sin x \,dx}{\sqrt[3]{3+2\cos x}}\]

Замена $y=\cos x$

\[ I=-\int\frac{dy}{(3+2y)^{\frac 13}}=-\frac 12\int(2y+3)^{-\frac 13}d
(2y+3)=-\frac 12(2y+3)^{\frac 23}\,\frac 32=-3(3+\cos x)^{\frac 23}\]

23.
\[ \int\frac{(x^2-x+1)\,dx}{x^4+2x^2-3}=\int\frac{(x^2-x+1)\,dx}
{(x-1)(x+1)(x^2+3)}\]

разлагаем дробь на три части

\[ \frac A{x-1}+\frac B{x+1}+\frac{Cx+D}{x^2+3}=
\frac{x^2-x+1}{(x-1)(x+1)(x^2+3)}\qquad
 \left\{\begin{array}{l}
A+B+C=0\\ A-B+D=1\\ 3A+3B-C=-1\\ 3A-3B-D=1\end{array}\right.\qquad
\begin{array}{l}
A=1/8\\ B=-3/8\\ C=1/4\\  D=1/2\end{array}\]

\[ I=\frac 18\left(\int\frac{dx}{x-1}-3\int\frac{dx}{x+1}+
\int\frac{2x\,dx}{x^2+3}+4\int\frac{dx}{x^2+3}\right)=
\frac 18\left(\log|x-1|-3\log|x+1|+\int\frac{d\,x^2}{x^2+3}+\right.\]
\[ \left.
\frac 4{\sqrt{3}}\int\frac{d\frac x{\sqrt{3}}}{\left(\frac x{\sqrt{3}}
\right)^2+1}\right)=
\frac 18\left(\log|x-1|-3\log|x+1|+\log(x^2+3)+\frac 4{\sqrt{3}}\arctg
\frac x{\sqrt{3}}\right)\]

24.
\[ \int\frac{(\sqrt[3]{4+\log x})\,dx}{x}\qquad
\mbox{замена} \quad \log x=y\]

\[ I=\int (4+y)^{\frac 13}\,dy=\frac 34(y+4)^{\frac 43}=
\frac 34(4+\log x)\sqrt[3]{4+\log x}\]

25.
\[
\int\frac{(x^-6)\,dx}{x^4+6x^2+8}=\int\frac{x^3-6}{(x^2+2)(x^2+4)}\,dx\]

\[
\frac{Ax+b}{x^2+2}+\frac{Cx+d}{x^2+4}=\frac{x^3-6}{(x^2+2)(x^2+4)}\qquad
\left\{\begin{array}{l} A+C=1\\ 4A+2C=0\\ B+D=0\\
4B+2D=-6\end{array}\right.\qquad \begin{array}{l} A=-1\\ B=-3\\ C=2\\
D=3\end{array}\]
\[ I=-\frac 12\int\frac{2x\,dx}{x^2+2}-3\int\frac{dx}{x^2+2}+
\int\frac{2x\,dx}{x^2+4}+3\int\frac{dx}{x^2+4}=
-\frac 12\log(x^2+2)+\log(x^2+4)-\frac 3{\sqrt{2}}\arctg\frac
x{\sqrt{2}}+\frac 32\arctg\frac x2\]

26.
\[ \int x\arcsin \frac 1x\,dx,\qquad
\mbox{берем по частям}\quad u=\arcsin\frac 1x,\quad v=x^2/2\]
\[I=\frac{x^2}2\arcsin\frac 1x-\int\frac{x^2}2\frac
1{\sqrt{1- \frac 1{x^2}}}\left(-\frac 1{x^2}\right)\,dx=
...+\frac 12\int\frac{x\,dx}{\sqrt{x^2-1}}=...+\frac 12\int
(x^2-1)^{-\frac 12}\,d\,x^2=\frac{x^2}2\arcsin\frac 1x+
\frac 12\sqrt{x^2-1}\]


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27.
\[ \int\frac{(\sqrt[4]{x}+1)\,dx}{(\sqrt{x}+4)\,\sqrt[4]{x^3}},\qquad
\mbox{замена}\quad \sqrt[4]x=y\]
\[ I=4\int\frac{y+1}{y^2+4}\,dy=2\int\frac{d\,y^2}{y^2+4}+
2\int\frac{d\frac y2}{\left(\frac y2\right)^2+1}=
2\log(y^2+4)+2\arctg\frac y2=
2\log(\sqrt{x}+4)+2\arctg\frac{\sqrt[4]{x}}2\]

28.
\[ \int x\log(x^2+1)\,dx,\qquad \mbox{замена}\quad x^2+1=y\]
\[ I=\frac 12\int\log y\,dy=\frac 12(y\log y-y)=
\frac 12(x^2+1)(\log(x^2+1)-1)\]

29.
\[ \int\frac{\sqrt{x+5}\,dx}{1+\sqrt[3]{x+5}},\qquad \mbox{замена}
\quad y=\sqrt[6]{x+5}\]
\[ I=6\int\frac{y^8\,dy}{1+y^2}=6\int\frac{y^8-1+1}{y^2+1}\,dy=
6\left(\int\frac{(y^2-1)(y^2+1)(y^4+1)}{y^2+1}\,dy+
\int\frac{dy}{1+y^2}\right)=\]\[
6\left(\int(y^6+y^2-y^4-1)\,dy+\arctg y\right)=
6\left(\frac {y^7}7+\frac{y^3}3-\frac{y^5}5-y+\arctg y\right)=\]\[
6\left(\frac 17(x+5)^{\frac 76}+\frac 13(x+5)^{\frac 12}-
\frac 15(x+5)^{\frac 56}-(x+5)^{\frac 16}+\arctg(x+5)^{\frac
16}\right)\]

30.
\[ \int x\sin x\cos x\,dx=\frac 12\int x\sin 2x\,dx,\qquad
\mbox{берем по частям}\quad u=x,\quad v=-\frac 12\cos 2x\]
\[ I=-\frac 14x\cos 2x+\frac 14\int\cos 2x\,dx=
-\frac 14x\cos 2x+\frac 18\sin 2x\]

31.
\[ \int x^2\sin 4x\,dx,\qquad \mbox{замена}\quad 4x=y,\qquad
I=\frac 1{64}\int y^2\sin y\,dy,\qquad\mbox{берем по частям}\quad
u=y^2,\quad v=-\cos y,\]
\[ I=\frac 1{64}\left(-y^2\cos y+\int\cos y\,2y\,dy\right),\qquad
\mbox{по частям}\quad u=y,\quad v=\sin y,\]
\[I=
\frac 1{64}\left(-y^2\cos y+2y\sin y-2\int\sin y\,dy\right)=
\frac 1{64}\left(...+2\cos y\right)=
\frac 1{32}\left(4x\sin 4x+(1-8x^2)\cos 4x\right)\]


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33.
\[ \int x\log^2x\,dx,\qquad\mbox{по частям}\quad u=\log^2x,\quad
v=\frac{x^2}2,\quad I=\frac{x^2\log^2x}2-\int\frac{x^2}2\,2\log
x\,\frac 1x\,dx=
...-\int x\log x\,dx\]
\[ \mbox{по частям}\quad u=\log x,\quad v=\frac{x^2}2,\quad
I=...-\frac{x^2}2\log x+\int\frac{x^2}2\,\frac 1x\,dx=...+\frac 14x^2=
\frac{x^2}2\left(\log^2 x-\log x+\frac 12\right)\]

34.
\[ \int\frac{dx}{\sqrt{x+3}+\sqrt[3]{(x+3)^2}},\qquad\mbox{замена}\quad
\sqrt[6]{x+3}=y\]
\[ I=\int\frac{6y^5\,dy}{y^3+y^4}=6\int\frac{y^2\,dy}{1+y}=
6\int\frac{y^1-1+1}{y+1}\,dy=6\left(\int(y-1)dy+\int\frac{dy}{y+1}\right)=
\]\[6\left(\frac{y^2}2-y+\log|y+1|\right)=
6\left(\frac{\sqrt[3]{x+3}}2-\sqrt[6]{x+3}+\log|\sqrt[6]{x+3}+1|\right)\]

35.
\[ \int\frac{(\sqrt{x}-1)(\sqrt[6]{x}+1)\,dx}{\sqrt[3]{x^2}},
\qquad\mbox{заменa}\quad \sqrt[6]{x}=y\]
\[ I=\int\frac{(y^3-1)(y+1)\,6y^5\,dy}{y^4}=
6\int\left(y^5+y^4-y^3-y\right)\,dy=
6\left(\frac{y^6}6+\frac{y^5}5-\frac{y^4}4-\frac{y^2}2\right)=
x+\frac 65x^{\frac 56}-\frac 32x^{\frac 23}-3x^{\frac 13}\]

36.
\[ \int\frac{\cos x\,dx}{\cos x+1}=\int\frac{\cos x+1-1}{\cos x+1}\,dx=
x-\int\frac{dx}{\cos x +1}=\]
\[ x-\int\frac{\cos x-1}{(\cos x+1)(\cos x-1)}\,dx=
x+\int\frac{\cos x-1}{\sin^2 x}\,dx=
x+\int\frac{\cos x\,dx}{\sin^2 x}-\int\frac{dx}{\sin^2 x}=\]
\[ x+\int\frac{d\sin x}{\sin^2x}+\cot x=x-\frac 1{\sin x}+\cot x\]

37.
\[ \int\frac{dx}{3\cos x+4\sin x}=\int\frac{3\cos x-4\sin x}
{9\cos^2x-16\sin^2x}\,dx=\]
\[3\int\frac{\cos x\,dx}{...}-4\int\frac{\sin x\,dx}{..}=
3\int\frac{d\,\sin x}{9-25\sin^2x}+4\int\frac{d\,\cos
x}{25\cos^2x-16}\]

в первом интеграле делаем замену $\sin x=y$

\[ I_1=\int\frac{dy}{9-25y^2}=\int\frac{dy}{(3-5y)(3+5y)}=
\frac 16\left(\int\frac{dy}{3-5y}+\int\frac{dy}{3+5y}\right)=\]\[
\frac 1{30}\left(-\log|5y-3|+\log|5y+3|\right)=
\frac 1{30}\log\left|\frac{5y+3}{5y-3}\right|=
\frac 1{30}\log\left|\frac{5\sin x+3}{5\sin x-3}\right|\]

во втором интеграле делаем замену $\cos x=y$

\[ I_2=\int\frac{dy}{25y-16}=\frac 18\left(\int\frac{dy}{5y-4}-
\int\frac{dy}{5y+4}\right)=\frac
1{40}\log\left|\frac{5y-4}{5y+4}\right|=
\frac 1{40}\log\left|\frac{5\cos x-4}{5\cos x+4}\right|\]

\[ I=\frac 1{10}\left(\log\left|\frac{5\sin x+3}{5\sin x-3}\right|+
\log\left|\frac{5\cos x-4}{5\cos x+4}\right|\right)=
\frac 1{10}\log\left|\frac{(5\sin x+3)(5\cos x-4)}{(5\sin x-3)
(5\cos x+4)}\right|\]

\end{document}
