Решить интегралы с подробным решением. Created by корчик. algebra-ru

Решить интегралы с подробным решением...

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Task/26882565
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7.
∫x*√(5x²+1)dx =(1/10)*∫√(5x²+1)d(5x²+1) =(1/10)*(2/3)*√(5x²+1)³ +C=
(1/15)*√(5x²+1)³ +C
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8.
∫(x+2) /(2x²+8x -7) dx =(1/4)*∫(1 / (2x²+8x -7) )* d(2x²+8x -7) =
(1/4)*Ln|2x²+8x -7| +C .
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9.
∫ x*Lg(x+2)*dx =(1/Ln10)*∫ x*Ln(x+2)*dx =(1/2Ln10)*∫ Ln(x+2)*d(x²) =
(1/2Ln10)*( x²*Ln(x+2)-∫ x²d(Ln(x+2)=(1/2Ln10)*( x²*Ln(x+2)-∫x²/(x+2)*dx )=
(1/2Ln10)*( x²*Ln(x+2) - x²/2 +2x - 4*Ln(x+2) ) + C .
* * * ∫ x²/(x+2) *dx = ∫ (x²- 4+4)/(x+2) *dx= ∫ (x-2)*dx + ∫ 4 /(x+2) *dx=
x²/2 -2x +4*Ln(x+2) * * *
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10.
∫ 1/ 9x²+25) +2/cos²(2x) )*dx =∫1/(9x²+25)*dx +∫2 /cos²(2x) *dx =
=(1/25)*∫(5/3)/(1 +(3x/5)²)*d(3x/5) +∫1/cos²(2x) *d(2x) =
(1/25)*(5/3)*∫1/(1 +(3x/5)²)*d(3x/5) +∫1/cos²(2x) *d(2x) =
(1/15)*arctg(3x/5) +tg(2x) +C .
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11.
∫sin²x*cosx*dx =∫sin²x*d(sinx) = (1/3)*sin³x +C.
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12.
∫cos(2x²+3)*x*dx =(1/4)*∫cos(2x²+3)*d(2x²+3) =(1/4)*sin(2x²+3)+C.

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$7)\; \; \int x\, \sqrt{5x^2+1}\, dx=[\, t=5x^2+1\; ,\; dt=10x\, dx\, ]=\\\\= \frac{1}{10}\cdot \int t^{1/2} \cdot dt= \frac{1}{10}\cdot \frac{t^{3/2}}{3/2}+C= \frac{1}{10}\cdot \frac{2\, \sqrt{(5x^2+1)^3}}{3}+C\\\\8)\; \; \int \frac{x+2}{2x^2+8x-7}\, dx=[\, t=2x^2+8x-7\; ,\; dt=(4x+8)dx,\\\\dt=4(x+2)dx\, ]=\frac{1}{4}\, \int \frac{dt}{t}= \frac{1}{4}\cdot ln|t|+C=\frac{1}{4}\cdot ln|2x^2+8x-7|+C$

$9)\; \; \int x\cdot lg(x+2)\, dx=[\, u=lg(x+2)\; ,du=\frac{dx}{(x+2)ln10}\, ,\\\\dv=x\, dx\; ,\; v= \frac{x^2}{2}\, ]=uv-\int v\, du=\\\\ =\frac{x^2}{2}\cdot ln(x+2)-\int \frac{x^2\, dx}{2(x+2)\, ln10}=\frac{x^2}{2}\cdot ln(x+2)-\\\\-\frac{1}{2\, ln10}\int (x-2+\frac{4}{x+2})dx=\frac{x^2}{2}\cdot ln(x+2)-\\\\-\frac{1}{2\, ln10}\cdot (\frac{x^2}{2}-2x+4\, ln|x+2|)+C$

$10)\; \; \int \Big (\frac{1}{9x^2+25}+ \frac{2}{cos^22x}\Big )dx= \int \frac{dx}{(3x)^2+5^2}+\int \frac{\overbrace {2\, dx}^{d(2x)}}{cos^22x} =\\\\= \frac{1}{5}\cdot \frac{1}{3}\cdot arctg \frac{3x}{5}+tg2x+C\\\\11)\; \; \int sin^2x\cdot cosx\, dx=[\, t=sinx\; ,\; dt=cosx\, dx\, ]=\\\\=\int t^2\, dt= \frac{t^3}{3}+C=\frac{sin^3x}{3}+C$

$12)\; \; \int cos(2x^2+3)\cdot x\, dx=[\, t=2x^2+3\; ,\; dt=4x\, dx\, ]=\\\\=\frac{1}{4}\cdot sint+C=\frac{1}{4}\cdot sin(2x^2+3)+C$

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