∫x+2x√3−6x32x  dx∫x+2x3−6x32x  dx = ∫3sinh(lnx2)  dx∫3sinh⁡(ln⁡x2)  dx =  ∫69+x2  dx∫69+x2  dx =

Q: 2x dx e2* +3e" +2 5.

A: To solve this integral, a substitution of is taken. Then it is differentiated to find the value of…

Q: 5- (tanx + secx)2 dx

A:

Q: (x-1) 12-3x dx

A: It can be solved directly

Q: sec® x dx cot x SE V

A: We can evaluate the given integral by using the substitution u= tan x and the following formulas.…

Q: 5x²+3 dx x (x²+2)

A: Given: Let us consider the given,∫5x2+3xx2+2dx

Q: 3 So 20 x² +9 dx

A:

Q: d/dx [(2x+1)3(3x2-1)4]

A:

Q: 6. 1 -dx x3 4

A: We have to find integration and we have to substitute limits later.

Q: 4sinx+sinxsecx dx cos²xV4+secx

A:

Q: d.x 2x2 +3x + 1 0. 2.

A:

Q: 1 √₁-² 1- Evaluate -dx (1 - 3x)²

A: “Since you have posted multiple questions, we will provide the solution only to the first question…

Q: x²+2x - 1 dx 2x3 + 3x2 - 2x

A:

Q: 4 dx x2 4x2 - -- x2

A:

Q: For what value(s) x does the graph of f(x) ==- x² + 3 have a horizontal tangent? For what value of…

A: For horizontal tangent first derivative must be zero

Q: 는(csc-1(324)) = d SC dx

A:

Q: e-) 1-2x - x2

A:

Q: da d²y +6- dx dx² + 11 dx+6y = ex + cosx бу

A:

Q: (5x+sin3(4x) + sin(4x³)) dx

A: use differentiation for finding the solution.

Q: Integrate

A:

Q: x + 5X" + 4X - 3 dx (x² + 3x + 5) (x²+7X+12)

A: The given integral is ∫x6+5x4+4x-3x2+3x+5x2+7x+12dx.

Q: Ssec? [sec(4x)tan(4x) dx (p sec (5x +3) dx e) f sin 3x + 1. dx 2x +4 f)

A: We will solve the following

Q: (5) Evaluate 75 2x -dx. 5x8

A:

Q: 5(3х-1) dx sec ( 3x²-2x)

A: On doing integration we will get