S (x5 –x³ – 6x) dx is the same as x²+2 A В (x³+3)dx 4x dx x2+2 D +x-- 8x dx x²+2 E F None

Q: y² = 2x = x² 1- y² = (x - 1)² x = 1+√1-y²

A: (a-b)2 = a2 - 2 a b +b2

Q: Var(5 + 2vx) = 4(E(x) – (E(Jx))² true False O

A: From the provided information, Var (5 + 2√x) = 4(E (x) – (E(√x)))2

Q: dx J (x² + 2x – 10)²

A:

Q: J(x-3x)dx = x' 3x +C 2 -3х + С 4 b. 3x-3+C 3x+C 4 4x* -6x +C

A:

Q: 2x + 1 (b) √ ² (1 − x) -dx

A:

Q: S (et +3e²x) dx 2x 5+4et-e²x

A:

Q: 5x3 + 6x2 + 2, f (0) 3 and f(1) -2. %3D 15. Find f for f"(x)

A:

Q: Find the antiderivative of  x4(5x5 + 1)5

A: To find out the antiderivative we use u- substitution method

Q: Differentiate the function below. y= (x2 + 4) (x² – 2) 8 d [(x? +4) (x² - 2) ® _ = O %3D dx

A: Ans is given below

Q: 2x - 2x2-3 dx - Find 4 x* + x²

A: To evaluate: ∫2x3−2x2−3x4+x2dx Let, I=∫2x3−2x2−3x4+x2dx ⇒I=∫2x3−2x2−3x2x2+1dx

Q: Find the value of ( x + y ) / (x - y ) if x = 6 and y = - 2.

A: The expression: x+yx-y

Q: 5x2-12 х-12 dx х3-4х

A: Given: ∫5x2-2x-12x3-4x

Q: 3x2 dx (x-1)(x2 + x + 1)

A:

Q: y = x+ – 4x³/2 (x > 0) (x>0)

A: Consider the function:

Q: ((-2x +5)dx

A:

Q: х3 (2 — х?)1?dx x3 dx 1 — 2х2 1 x²dx Vx2 + 4

A: ∫un du = un+1n+1+C ∫a2+x2 dx = xx2+a22+a22sin h-1x2+C ∫1x2+a2 dx = sin h-1xa+C

Q: Find f(₂ + 2)(x - 7)dx + C

A: The given integral is: I=∫x+2x-7dx   As we know that, The Sum Rule: ∫fx±gxdx=∫fxdx±∫gxdx The Power…

Q: x² dx x²-3x+2

A:

Q: x-1 2 x² + x -dx

A: As per the question we are given an indefinite integral of (x-1)/(x2 + x) which we have to evaluate.

Q: dx x2 – 2x + 2

A:

Q: x-4 -dx x² – 5x +6

A:

Q: 3x2 + 3x - 1 dx. (x – 1)(x² + 2x + 2)

A: In the given question we have to calculate the integral ∫3x2+3x-1(x-1)(x2+2x+2)dx. i will directly…

Q: x²-3x - 7 √(x² + x + 2) (2x - 1) dx

A:

Q: (2х — 8) dx 2 х — 8х + 25 X

A:

Q: 2sin(2x)=√3

A:

Q: d -(f(x² + 3x)) dx

A: Derivative

Q: | (7x + 4) e dx

A:

Q: 4a+ax/x)=f a=

A:

Q: 3x-6 dx VE-2

A:

Q: x-3 dx (2x–10x+13) /10x²–42x+45

A: To evaluate the given indefinite integral.

Q: Find the value of x

A:

Q: 10x3 dx 1+ x³

A: Given ∫10x31+x3dx

Q: Evaluate V1-x-dx A. --(x+xvI– x-) +C B. (x+xV1%3=x) +C c. (x+xv1+x-) +C D. -(x-xVT-x-) + C xv1-x)+ C

A: Find your answer below