dy Given y= x³In x, -is dx A 3x2In x B B 3x2in x+ x2 ( c) x2 D 3x2

Q: Determine  d2y dx2 for y = (4x3 + 8)4.

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Q: 3 2x dx 2? - 4х + 5 C. 2)

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Q: x1 J: (x2+2x)2 dx= (x2+2x)3 C (x2 2x)3 C + 3 (x2+2x)3 +C 6 22-1C + -2 2x)1+C

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Q: 2x+3 a) S; dx x2-3x+2

A: We integrate the following by using substitution rules and factories the denominator.

Q: dx V3+2x-x2 10.

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Q: Evaluate (x² + y²) dxdy

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Q: (a1x + bịy + c1)dx + (a2x + b2y + c2)dy = 0

A: We will find out the required general solution.

Q: Find d²y dx² V = d²y 12 || 9,3 3 – 5X -

A: .

Q: f. S r4 dx J2 xV2-2x2 2.

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Q: dx Vx? + 2x + 5

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Q: X2 dx ex+3

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Q: 5 4x³-3x²+12 10. Find y = dx

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Q: dx x2 – 2x + 2

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Q: 5. (2x + 3V )dx 2х + 3уx

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Q: Refer to the graph shown below and evaluate f(t)dt. f(t)dt = (Simplify your answer.) -5 - 5 5+t 5-t

A: Given:                ∫-55ftdt  From the above graph            f(t)= 5+t            f(t)=5-t

Q: (x2-3) Cx²+ 2) S dx J3xx3

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Q: sx2-9x-6 dx x3-2x2- 4xt8

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Q: d²y if 2x²-3y²=7 dx2

A: Differentiating both sides of the equation. 23x2-32ydydx=06x2-6ydydx=06ydydx=6x2dydx=x2y . . . . . .…

Q: f 12x -9x +2 dx 3.

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Q: (x3-4) 2. S dx x3+2x2+2x

A: To evalaute: I=∫x3-4x3+2x2+2xdx=∫1-2x2+2x+4x3+2x2+2xdx=∫dx-∫2x2+2x+4x3+2x2+2xdx

Q: (2x - y)dx(x + 4y)dy = 0

A: As per the rules, we can answer only the first question and other can be reposted.

Q: 2.

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Q: dx 2. Evaluate -2 v2x2 + 8x + 10

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Q: dx Vx4 – 2x² + 1 3,

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