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e^(3x^2-8 * 6x 6xe^(3x-8)
5ln(12x^2-20) + ln(25x^2-2x) 5 * (1/12x^2-20) * 24x + 1/(25x^2-2x) * 50x - 2 (120x) / (12x^2-20) + (50x-2)/25x^2-2x
definite integral
integral (2,0) 2x+1 dx 2 is upper 0 is lower on the symbol 2x^2/2 + 1x x^2 + x
plug in 2 minus plug in 0
2^2 + 2 = 6 0^2 + 0 = 0 6 - 0 = 6 answer is 6
e^(-.2x^2 + 4) * -.4x = 0 zero at zero because -.4(0) = 0
e^(-.2x^2+4) = 0 Never will happen. horizontal asymptote at x = 0. ln(0) is undefined and ln is the reverse of e^.
so we only have 0
so now solve at -1, 0, and 1
-.4(-1)e^(-.2(-1)^2 + 4) .4e^(-.2(1) + 4) .4e^(-.2 + 4) .4e^(3.8) positive
-.4(1)e^(-.2(1)^2+4) -.4e^(-.2+4) -.4e^(3.8) negative
f increasing -inf, 0 f decreasing 0,inf
-8x^8 + 9/x - 7/x
(-8x^9)/9) + 9ln(|x|) - 7(x^-8/-8) + c -8/9 x^9 + 9ln(|x|) - (7x^-8)/-8 + c -8/9x^9 + 9ln(|x|) - 7/8x^8 + c
u substitution caues we have inside funtcion
inside exponent denominator
inside
u = (7x^4+4x-3) du = (28x^3 + 4)dx now we do u * du 28x^3 + 4 already matches up integral u^7 du u^8/8 + c ((7x^4+4x-3)^8)/8 +c
denominator u substitution 5x+5 is u
u = 5x+5 du = 5dx
but we have no multiple by 5
its just dx
so we multiply whole thing by 5
1/5 integral(9,6) 1/(5x+5) * 5dx we have 1/5 int(9,6) 1/u 1/5 ln|u| 1/5 ln|5x+5| integral(9,6) plug in 9 - plug in 6
1/5ln 5*9+5 1/5ln(50) - 1/5ln(35) 1/5ln(50/35)
this means find c too
-5e^x - 5x
F(0) = 6
integral(-5e^x - 5x) -5e^x - 5x^2/2 -5e^x - 2.5x^2 + c
-5e^0 - 2.5(0)^2 + c = 6 -5e^0 + c = 6 -5 + c = 6 c = 6+5 c = 11
-5e^x - 2.5x^2 + 11
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