3 + 2х +1 2х +1 2 dx -4

Q: 1 -dx 3-4х

A:

Q: 3. | (1- 2x) dx 0.

A: Given that, ∫01xdx=12 ∫01x2dx=13 ∫01x3dx=14 The integral to be evaluated is given as, I=∫011-2x3dx…

Q: 3x+2 2+9x+1 dx

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Q: 6. (4x-6y- 1) dx + (3y – 2x – 2) dy = 0.

A:

Q: -2 (5х2-3х+18)dx 1. 1 9х-х3

A: We use partial fraction to find the given integral As per our guideline we supposed to answer only…

Q: 2х — 3 - dx (х-1)(х-2)(х+ 3)

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Q: бх+8 dx (х+3)(х+5)(х-2)

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Q: 3x 1. 31. -dx X +4 2. 2x3 2 3-4 dx

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Q: 4- I = 57 4x+1 dx

A: Given: I=∫74x+1dx For evaluating given integral, we simplify given expression then integrate it

Q: 2 2 4 dx%3D 3 = - || %3D -1 x' -1 3/2

A:

Q: intergral 1/20 1/((x-1/2)2 + 3/4) dx

A: Formula Used: ∫1x2+1dx=tan-1 (u)tan-(0)=0tan-1-13=-π6

Q: (3х+5) dx х3-х2-х+1

A: Using partial fractions we will split the given integral,then we will apply standard integral…

Q: (х - 2у + 3) dx + (3х — бу + 1) dy %3D о

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Q: 5. 2x-1 4x - 1 dx

A: We can solve this as follows:

Q: -4 6. |a + 3|dx – 1)³ydy -1

A:

Q: 2x-4 | 5e dx

A: To evaluate the given integral.

Q: 1 dx. -1 2x+3 -1

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Q: |sec-(1.8x – 2.3) dx = %3D

A: Let the given integral is I=∫ sec21.8x-2.3dx By using the substitution, Put, u = 1.8x - 2.3…

Q: ニ 2 2. dx | -r

A: The given integral  y=2∫-rrr2-x2dx  Can be solved using integration formulas given below…

Q: 2х + 1 бх2 + 23х — 9 dx х3 + 2х2 — 3х Ja- 2)(x – 3)² dx

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Q: .3, 2- 2 1. dx =? 4

A: ∫02x2-x342dx = ∫02x2-x616dx

Q: 2 (3x- 2) dx Вх — -2

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Q: 1. ƒ'¸ (3 - 2x)³ dx

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Q: (2х - 9)4 dx

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Q: 24 - dx /3- 4x (х + 1) 3.

A: Given that:

Q: 3 +5 - dx =

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Q: 6. 3(у + 1) dx %3D 2х dy

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Q: /6- х2 dx

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Q: 6. - 2х + 3) dx

A: According to Bartleby's guidelines we can solve only one question at a time. Apologise for the…

Q: (2² –4)* -dx (x–2)²

A: First we simplify numerator with using appropriate identities then integrate

Q: 2х—3 1./ —1x2 — Зх + 1 dx

A: Given, the integral

Q: /9-4х2 dx

A: We know, (i) ∫dxx2-a2=12alogx-ax+a+C, C is a constant.

Q: (4x - 3)2/3 dx

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Q: 2 Væ2 – 3 dx V3

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Q: -dx 23+1

A: Given :

Q: ∫〖7(4x-1)^6 dx〗

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Q: 4 - 1.5x +1.3 – x 3 dx

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Q: 3 - 2 (6х + у) dx dy о -5 3 - 2 (6х + у) dx dy 3

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Q: (1-2х)2 dx 3.

A: Given :       ∫1 - 2x2x dx   To evaluate the given integral.

Q: 5(3х-1) dx sec ( 3x²-2x)

A: On doing integration we will get

Q: 3. f(2x + 3)² dx

A: I have stated the used formulas in the solution

Q: 0 1 S -8 (2x-1)3 -dx

A: ∫-∞0 12x-13dx

Q: 2. 3. 4 了- dx

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Q: 2х+1 dx (3х-1)(х2+2х+2)

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Q: 2 (2х + Зу + 5) dx) dy -1 У-2

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Q: · 1. -1 (3-2x) ³ dx

A: The objective is to evaluate the integration.

Q: (3х+5)dx х3-х2-х+1

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Q: 2x +1 -dx J - 3x +2` 4. х — Зх

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Q: 4x2 - 3х + 2 dx. Evaluate 4x - 4х +3

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Q: .2 4- dx 2 (. 2)4 3.

A: