Интеграли, помощ !

[MuKCePaBG]

Здравейте,

Искам да Ви помоля за помощ за решаването на следните интеграли :


Благодаря много на хората, склонни да помогнат

Весела коледа и весели празиници !

[kmitov]

Не ти ли казаха какви числа да сложиш наместо буквите $a$ и $b$?

[MuKCePaBG]

Да!

а=3
b=0

[Nathi123]

б) [tex]I= \int tg^{2}4xdx=\int \frac{sin^{2}4x}{cos^{2}4x}dx=\int \frac{1-cos^{2}4x}{cos^{2}4x}dx=\int \frac{dx}{cos^{2}4x}-\int dx =\frac{1}{4}\int\frac{d4x}{cos^{2}4x}-x+C[/tex]
[tex]I=\frac{tg4x}{4}-x+C_{1 }[/tex].

[Nathi123]

г) [tex]I=\int \frac{8.e^{2x}dx}{1+e^{2x}}= 4\int \frac{d(1+e^{2x}}{1+e^{2x}} =4ln(1+e^{2x}) + C .[/tex]

[Nathi123]

в) [tex]I= \int\frac{dx}{\sqrt{x}(5\sqrt{x}+2)} ; \sqrt{x}=t; t\ge 0 \Rightarrow x=t^{2}\Rightarrow dx=2tdt\Rightarrow I=\int\frac{2tdt}{t(5t+2)}[/tex]
[tex]I=2\int \frac{dt}{5t+2}=\frac{2}{5}ln(5t+2)+C=\frac{2}{5}ln(5\sqrt{x}+2) + C.[/tex]

[aifC]

и) [tex]\int \frac{x-4}{\sqrt{x^{2}-8x+17}}dx; u=x^{2}-8x+17 \rightarrow dx = \frac{1}{2x-8}du[/tex]

[tex]\frac{1}{2} \int \frac{1}{\sqrt{u}}du = \sqrt{u} = \sqrt{x^{2}-8x+17} +C ;[/tex]

й) [tex]\int \frac{2sin(x)-5}{2cos(x)+5}dx = \int \left(\frac{2sin(x)}{2cos(x)+5} - \frac{5}{2cos(x)+5}\right)dx = 2\int \frac{2sin(x)}{2cos(x)+5}dx -5\int \frac{5}{2cos(x)+5}dx = I_{1} - I_{2};[/tex]

[tex]I_{1}= \int \frac{2sin(x)}{2cos(x)+5}dx; u=2cos(x)+5 \rightarrow dx = \frac{1}{2sin(x)}du \Rightarrow -\frac{1}{2} \int \frac{1}{u}du = -\frac{ln(u)}{2} = -\frac{ln(2cos(x)+5)}{2} + C;[/tex]

[tex]I_{2} = \int \frac{5}{2cos(x)+5}dx = \int \frac{sec^{2} \left(\frac{x}{2}\right)}{3tg \left(\frac{x}{2}\right)+7}dx; u = \frac{\sqrt{3}tg \left(\frac{x}{2}\right)}{\sqrt{7}} \rightarrow dx = \frac{2\sqrt{7}}{\sqrt{3}sec^{2}\left(\frac{x}{2}\right)}du;[/tex]

[tex]\frac{2}{\sqrt{3} \cdot \sqrt{7}}\int \frac{1}{u^{2}+1}du = arctg(u) = \frac{2arctg \left(\frac{\sqrt{3}tg \left(\frac{x}{2}\right)}{\sqrt{7}}\right)}{\sqrt{3}\cdot\sqrt{7}}+C;[/tex]

[tex]I_{1} - I_{2} = ln(2cos(x)+5) + \frac{10arctg \left(\frac{\sqrt{3}tg \left(\frac{x}{2}\right)}{\sqrt{7}}\right)}{\sqrt{3}\cdot\sqrt{7}}+C;[/tex]

[kmitov]

И тези решения са ала бала.

[MuKCePaBG]

Какво ще рече, че са "ала бала ?"
Евалата на хората, че са склонни да помогнат
Благодарско !!!

[MuKCePaBG]

Ако някой може да помогне с е), ж),з) че закъсах до там???

[aifC]

e) [tex]\int (6x^{5}+1)ln(x^{6}+x)dx; u = x^{6}+x \rightarrow dx = \frac{1}{6x^{5}+1}du;[/tex]

[tex]\int ln(u)du = uln(u) - \int 1du = u; \Rightarrow x(x^{5}+1)(ln(|x||x^{5}+1|)-1) +C;[/tex]

[aifC]

ж) [tex]\int(x^{3}+x)sin(4x)dx = \int x(x^{2}+1)sin(4x)dx = - \int -\frac{(3x^{2}+1)cos(4x)}{4} dx - \frac{x(x^{2}+1)cos(4x)}{4} = -\frac{1}{4} \int (3x^{2}+1)cos(4x)dx = \frac{(3x^{2}+1)sin(4x)}{4} - \int \frac{3x \cdot sin(4x)}{2}dx = \frac{3}{2} \int x \cdot sin(4x)dx = - \int - \frac{cos(4x)}{4} dx - \frac{xcos(4x)}{4}; u = 4x \rightarrow dx = \frac{1}{4}du \Rightarrow -\frac{1}{16} \int cos(u)du = sin(u)+C = \frac{(24x^{2}+5)sin(4x)+(-32x^{3}-20x)cos(4x)}{128} + C;[/tex]

[aifC]

з) [tex]\int \frac{5x^{2}-x+3}{(x^{2}+1)^{2}(x-5)}dx = \int \left(-\frac{123x+615}{676(x^{2}+1)} + \frac{7x+9}{26(x^{2}+1)^{2}} + \frac{123}{676(x-5)} \right)dx[/tex]

[tex]-\frac{123}{676} \int \frac{x+5}{x^{2}+1}dx + \frac{1}{26} \int \frac{7x+9}{(x^{2}+1)^{2}}dx + \frac{123}{676} \int \frac{1}{x-5}dx = -I_{1}+I_{2}+I_{3};[/tex]

[tex]I_{1} = \int \frac{x+5}{x^{2}+1}dx = \int \frac{x}{x^{2}+1} + 5 \int \frac{1}{x^{2}+1}dx = \frac{ln(x^{2}+1)}{2} + 5arctg(x)+C;[/tex]

[tex]I_{2} = \int \frac{7x+9}{(x^{2}+1)^{2}}dx = 7 \int \frac{x}{(x^{2}+1)^{2}}dx + 9 \int \frac{1}{(x^{2}+1)^{2}}dx = \frac{9arctg(x)}{2}+\frac{9x}{2(x^{2}+1)}-\frac{7}{2(x^{2}+1)} +C;

[tex]I_{3} = \int \frac{1}{x-5}dx = ln(x-5)+C \Rightarrow -I_{1}+I_{2}+I_{3} = \frac{2(123ln(|x-5|)-498arctg(x))-123ln(x^{2}+1)}{1352} + \frac{9x-7}{52x^{2}+52} + C;[/tex]

[kmitov]

Калпави решения. За 2.

[aifC]

So what