dx Use a trig substitution to evaluate x³ Vx² x > 1. 1 6. -

Q: Evaluate Sdx dx where A = 5x²-3x+2 and B = x³ - 2x². B

A: Evaluatedxwhere A = 5x2-3x+2B = x3-2x2i.e.

Q: exp(iz) = 1+ iV3.

A:

Q: Evaluate∫cos 6x dx without using the substitution method.

A:

Q: Differentiate y = 3e2x/(1 + x2.)

A: y=3e2x1+x2

Q: 1. 5x²+4x2-5 dx

A:

Q: Find - 6x+1 x² +1 -dx.

A:

Q: 7. Use substitution to evaluate ſ 2x(4 – x²) dx.

A:

Q: Evaluate f(4 – 3x)³ dr

A: The integration of the (4-3x)3 and limit from 1 to 2.

Q: x2+1 - dx x4-x²+1

A:

Q: dx Vx + a + Vx + b

A: Given: The integral is ∫xx+a+x+bdx

Q: 2 dx 8x + 9

A: To find the value of given integration  We use some basic results of integration and substitution

Q: Evaluate I = / a*. g³ Vx² – 3 dx

A:

Q: Evaluate /x* V5 - x*dx

A: We have to evaluate the integral: ∫x35-x2 dx Rewriting the integral, ∫x35-x2 dx=∫x2x5-x2 dx Now…

Q: Integrate x sqrt of (x+4) dx using algebraic substitution

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Q: x2 8. dx V9 – x2

A:

Q: Differentiate and simplify. f(x)=Je* +In (x² +4)

A:

Q: 1 10. dx x² Vx² – 4 -

A:

Q: Evaluate x(x2 + 1)3 dx.

A: Given ∫x(x2+1)3dx

Q: Evaluate -3)x+3 dx

A:

Q: 4 dx V1- 25x2

A:

Q: Evaluate dx. x2-4x + 8

A: Given, ∫xx2-4x+8dx  We have to evaluate it.

Q: ifferentiate the funtion y = 2 csc (1-3x)

A: Giveny=2cosec(1-3x)Aim: dydx

Q: Integrate using u-substitution or 2 guys in Jail. fx sin(2x²) dx 5. (4x³-2)7 (24x²)dx

A: (4)   I=∫xsin(2x2)dx∫sinxdx=-cosx+cdxndx=nxn-1

Q: dx z/L1(zx — 6) вх І

A: The given integration is: ∫x149-x2172dx

Q: || д хр -XS - 7

A: NOTE: Refresh your page if you can't see any equations. .

Q: 4x dx (1-x²)³

A:

Q: dx 12 · S V5 – 4x² + 4x

A: