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(x^3 + 2)^4 4(x^3+2)^3(3x^2)
u
du
u = x^2 = 1 du = 2xdc
u^3 / 3 + c
((x^2+1)^3)/3 + c
inside exponent denominator
u = x^3 + 1 du = 3x
sqrt(u) du u^(1/2) du u^3/2 / 3/2 + c 2/3u^(3/2) + c 2/3(x^3+1)^(3/2) + c
2t / (t^2+1)
u = t^2+1 du = 2tdt
1/t^2+1 * 2tdt
1/u du ln(|u|) + c ln(|t^2+1|) + c
integral : (7x^7 + 6x + 3)^4 (49x^6 + 6) dx what function when we take the derivative will give me this for u sub our 3 main strategies: inside exponent denominator
u = 7x^7 + 6x + 3 du means differential of u du = 49x^6 + 6 dx ( this is not just from up there, that is what happens when we differentiate u )
now write in terms of u and du
integral : u
u^5/5 + c (7x^7 + 6x + 3)^5 / 5 + c to check, take derivative of answer
1/5 ( 7x^7 +6x + 3)^5 +c
(1/5)(5)(7x^7+6x+3)^4
(7x^7 + 6x + 3)^4
u
du
we dont have an inside function, or an exponential function, but we do have a denominator so u is t^2 + 1
u = t^2 + 1
du is differential
du = 2t (dt)
1/(t^2+1) * 2t(dt)
1/u du = ln|u| + c ln|t^2+1| + c
inside function u = 3-4x^2 du = -8x
u^2 * du (3-4x^2)^2 * -8x(dx) ((3-4x^2)^3)/3
(3-4x^2)^2 * (-8/-8)xdx (-1/8) (3-4x^2)^2 * -8xdx
(-1/8)((3-4x^2)^3)/3 + c (-1/24)(3-4x^2)^3 + c
u = x^2 + 1 du = 2x(dx) trying to get to u^2 * du (1/2) (2x)(x^2+1)^2dx (1/2) (x^2+1)^2(2x)(dx) so now u^3/3 + c (1/2) (x^2+1)^3 /3 + c (1/6) (x^2+1)^3 + c u8
exponent
u = x^2 + 3 du = 2xdx
7 * integral ( e^(x^2+3) * xdx) 7/2 * integral ( e^(x^2+3) * 2xdx) 7/2 e^u + c
7/2e^(x^2+3) + c
integral : 5e^3x dx
u = 3x du = 3
5 * integral : e^3x * dx (5)(1/3) * integral: e^3x * 3dx (5/3) * integral: e^3x * 3dx
5/3e^(3x) + c
inside exponent denominator
inside is -3t-2
u = (-3t-2) du = -3
so
5(-3t-2)
(-3t-2)^-4 * 5dt -(5/3) integral: (-3t-2)^-4 * -3dt
u^-4 * du u^-3 / -3 * du (-5/3)((-3t-2)^-3 / -3) + C (-5/3)(-1/3)(3t-2)^-3 + c (5/9)(3t-2)^-3 + c
5/((9)(3t-2)^3) + c
6e
inside exponent denominator exponent
u = 7y du = 7
6e^7y dy e^7y * 6dy
6/7 * integral: e^7y * 7dy 6/7 * e^7y + c
1/(x+5)
inside exponent denominator denominator x+5
u = x+5 du = 1
1/u
means it is
ln|u| + c
ln|x+5| + c
inside exponent denonminator inside
u = x-8 du = 1
all good on the 1dx
(x-8)^-4 / -4 + c 1/(x-8)^4 * -(1/4) +c -1/4(x-8)^4 + c
5t^2 * e^(5t^3) inside exponent denominator denominator u = 5t^3 du = 15t
problem is e^(5t^3) * 5t
1/3 integral: e^(5t^3) * 15t
1/3 u + c 1/3 e^(5t^3) + c
e^(3x^5-7) * 15x^4 u = 3x^5 - 7 du = 15x
e^(3x^5-7) + c
inside exponent denominator denominator real problem is: 1/(1+.5x) * 350dx
u = 1 + .5x du = .5
700 * integral: 1/(1+.5x) * .5x 1/u so 700ln|u| + c 700ln|1+.5x| + c
and then other term is -15.6 so thats -15.6x
700ln|1+.5x| - 15.6x + c
700ln|1+.5(0)| - 15.6(0) + c = -70 700ln|1| +c = -70 0 + c = -70 c = - 70
700ln|1+.5x| - 15.6x - 70
5e^(4y) ( 5+4e^4y)^3 dy
inside exponent denominator
inside function
u = 5+4e^4y du = 4e^4y * 4 du = 16e
5/16 * integral: ( 5+4e^4y)^3 16e
5/16 u^3 5/16 u^4/4 + c 5/16 * 1/4 * u^4 + c 5/64u^4 + c
5/64 * (5+4e^4y)^4 + c
inside exponent denominator inside u = 3x^7 + 4x - 6 du = 21x^6 + 4
(3x^7 + 4x - 6)^8 / 8
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