O Integrate In 2x dx by using integration by part.

Q: Evaluate the integral. x sec (3x) dx + C

A: Topic = Integration

Q: Evaluate f 5z" In8x dz using integration by parts.

A: Given integration: ∫5x7 ln8x dx Use the integration by parts: ∫u v dx=u∫v dx-∫dudx∫v dx dx Here…

Q: Evaluate using integration by parts o a) (In 3x dx

A: As per the rules of the company we can only the first question and rest others can be reposted.

Q: dx T² + 4 2

A: We have to find value of indefinite Integration.

Q: fx arccos x dx 2.

A:

Q: evaluate using Integration by Parts.

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Q: +1 dx. (x+3:2-5)3

A: To evaluate  ∫x2+1x3+3x-53dx

Q: 5x2 dx et

A:

Q: Evaluate using integration by parts: fx sec-¹x dx

A: SOLUTION-

Q: Evaluate the integral. 3* dx V3* + 4

A:

Q: 4 (1+2 In x)* 6. J dx б.

A: We have to find

Q: In integrating the function fxIn Sx dr using integration by parts, what would be your du? du = In 5x…

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Q: Evaluate the integral. x sec (9x) dx +C

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Q: Evaluate the integral. бх + 2 dx = | x2 + 2x – 3 -

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Q: 1. fx³ In x dx

A: Evaluate the integral equation by integration by parts .

Q: Evaluate the inde integration, SX²J9+x' o

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Q: Evaluate the integral. 2x´dx 4 5.

A: ∫452x2dx

Q: (b) Use Integration by parts to show that ·4 In x ~dx= =-+2ln.x) +C

A:

Q: Without integrating, explain why: S, ?)°dx = o x(1+ x* -2

A: To show:     ∫-22x1+x22dx = 0 (without integrating)

Q: Find the Integration of 1 dx. (6x+7)6

A: To find: The integral of ∫dx6x+76. Formula used: Integration formula: Power rule of integration:…

Q: evaluate using Integration by Parts.

A: To find the given integral by using integration by part formula.

Q: Use U-substitution to find the integral: 2 1 dx

A: Given,          ∫1x22-1xdx

Q: 5e* dx d. J J1-e2x 1+cos (2x) е. sin2 (2x)

A: Using the substitution method and solve the integration explanation is given below...

Q: Evaluate f x³e2x dx using integration by parts

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Q: Use integration by parts to find S x² In 5x dx

A:

Q: Evaluate using integration by parts. 1)x2 In 6x dx

A: We are asked  to evaluate the integral

Q: 4 dx X-X-2Ą

A:

Q: Evaluate the integral: 5x sec (x) dx +C

A: Recall: ∫u.v dx = u∫v dx -∫dudx∫v dxdx   ( Product rule of integration) ∫ sec2x dx = tan(x)+C…

Q: Evaluate the integral. (Use C for the constant of integration.) 17xe2x dx (1 + 2x)?

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Q: To evaluate S (In x)² -dx, apply integration by parts twice. For the second application of…

A: Follow the procedure given below.

Q: What is the result of the following ? integration dx Vx 4

A:

Q: sec? (4x)(3 – tan(4x))³dx

A: Given integral is,  ∫sec24x3-tan4x3dx. Substitute t=3-tan4x.…

Q: use integration by parts to write the integral sec^7 (3x) dx in terms of integral sec^5 (3x) dx

A: ∫sec7(3x)dxPut 3x=uso 3dx=du=13∫sec7uduNow ∫sec7udu=∫sec5usec2udu

Q: Evaluate 4xcos 3x dx by using integration by part.

A:

Q: What change of variables would you use for the integral  ∫(4 - 7x)-6 dx?

A: Consider the following integral: I=∫4-7x-6dx Substitute 4-7x=t→-7dx=dt→dx=-17dt in I:…

Q: Evaluate the integral.

A: It is given that, ∫021x+33dxWe have to evaluate it.

Q: Evaluate the integral. dx g4 Vx² + 3

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Q: Use integration by parts to evaluate the defini 2xe dx = 1.5

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Q: 5x -dx (? + 1)(3x – 2)

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Q: dx 1+e2*

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Q: Evaluate the integral. x sec (8x) dx + C

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Q: evaluate using Integration by Parts.

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Q: Use the substitution x = 5 sec t to evaluate the integral / dx xVx2-25

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Q: Evaluate the integral. 2x2 4 dx = V2 6x

A: Answer

Q: 2. fx Arctan xdx

A:

Q: Use Integration by Algebraic Substitution 3 dx x +1

A:

Q: 3+3x dx

A:

Q: (a) In(4x) dx

A: Integration by parts:  ∫f(x)g′(x)dx = f(x)g(x) −∫f′(x)g(x)dx

Q: 16 dx Vx-1 6,

A: I=∫9161x-1dxPut u=x-1differentiate w.r.t.xdu=12xdx2xdu=dx2(u+1)du=dxas x→9  ,u→2 as x→16 ,…

Riemann Sum

Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus. 

Riemann Integral

Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.