Question 1

Find an anti derivative (or integral) of the following functions by the method of inspection.
1. sin 2x
2. cos 3x
3. e2x
4. (ax + b)2
5. sin 2x – 4 e3x

Accepted Answer

1. ∫ sin 2x dx = -1/2 cos 2x + C

2. ∫ cos 3x dx = 1/3 sin 3x + C

3. ∫ e2x dx = 1/2 e2x + C

4. ∫ (ax + b)^2 dx = ∫ (a^2 x^2 + 2abx + b^2) dx = a^2 ∫ x^2 dx + 2ab ∫ x dx + b^2 ∫ dx = a^2 (x^3/3) + 2ab (x^2/2) + b^2 x + C = (a^2 x^3)/3 + ab x^2 + b^2 x + C

5. ∫ (sin 2x – 4 e3x) dx = ∫ sin 2x dx – 4 ∫ e3x dx = (-1/2) cos 2x – 4 (1/3) e3x + C = (-1/2) cos 2x – (4/3) e3x + C — Use basic integration formulas and linearity of integration:

1. ∫ sin kx dx = -1/k cos kx + C
2. ∫ cos kx dx = 1/k sin kx + C
3. ∫ e^(kx) dx = 1/k e^(kx) + C
4. Expand (ax + b)^2 and integrate termwise.
5. Use linearity: integrate each term separately.

Question 2

∫(4 e^{3x} + 1) dx

Accepted Answer

To evaluate ∫(4 e^{3x} + 1) dx, split the integral:

∫4 e^{3x} dx + ∫1 dx

First integral:
∫4 e^{3x} dx = 4 ∫ e^{3x} dx
Using substitution u = 3x, du = 3 dx => dx = du/3
So, ∫ e^{3x} dx = (1/3) e^{3x} + C
Therefore, ∫4 e^{3x} dx = 4 * (1/3) e^{3x} = (4/3) e^{3x}

Second integral:
∫1 dx = x + C

Combining:
∫(4 e^{3x} + 1) dx = (4/3) e^{3x} + x + C — Split the integral into two parts and integrate each separately. Use substitution for the exponential term and direct integration for the constant term.

Question 3

∫ x^2 (1 – 1/x^2) dx

Accepted Answer

Rewrite the integrand:

x^2 (1 – 1/x^2) = x^2 – 1

So, ∫ x^2 (1 – 1/x^2) dx = ∫ (x^2 – 1) dx = ∫ x^2 dx – ∫ 1 dx

Integrate each term:
∫ x^2 dx = (x^3)/3 + C
∫ 1 dx = x + C

Therefore,
∫ x^2 (1 – 1/x^2) dx = (x^3)/3 – x + C — Simplify the integrand by distributing and then integrate term by term.

Question 4

∫ (a x^2 + b x + c) dx

Accepted Answer

Integrate each term separately:

∫ a x^2 dx = a ∫ x^2 dx = a (x^3 / 3) + C
∫ b x dx = b ∫ x dx = b (x^2 / 2) + C
∫ c dx = c x + C

Therefore,
∫ (a x^2 + b x + c) dx = (a x^3)/3 + (b x^2)/2 + c x + C — Use linearity of integration and power rule for each term.

Question 5

∫ (2 x^2 + e^x) dx

Accepted Answer

Split the integral:

∫ 2 x^2 dx + ∫ e^x dx

First integral:
∫ 2 x^2 dx = 2 ∫ x^2 dx = 2 (x^3 / 3) = (2/3) x^3

Second integral:
∫ e^x dx = e^x

Therefore,
∫ (2 x^2 + e^x) dx = (2/3) x^3 + e^x + C — Integrate polynomial term using power rule and exponential term directly.

Question 6

∫ \left( x - \frac{1}{x} \right) dx

Accepted Answer

Rewrite the integral:

∫ (x - 1/x) dx = ∫ x dx - ∫ (1/x) dx

Integrate each term:
∫ x dx = x^2 / 2
∫ (1/x) dx = ln|x|

Therefore,
∫ \left( x - \frac{1}{x} \right) dx = \frac{x^2}{2} - \ln|x| + C — Split the integral and use power rule and logarithmic integration.

Question 7

∫ \frac{dx}{x^3 + 3x + 4}

Accepted Answer

This integral involves a rational function with a cubic denominator. Since the denominator is not factorized easily, the integral may require partial fraction decomposition or substitution if possible. However, as it stands, it is a standard integral requiring factorization or special methods.

Since the problem is from an exercise, the expected approach is to attempt factorization or use substitution if possible.

Step 1: Try to factorize denominator x^3 + 3x + 4.
Try rational roots: ±1, ±2, ±4

Test x = -1:
(-1)^3 + 3(-1) + 4 = -1 -3 +4 = 0

So, x = -1 is a root.

Therefore, denominator factors as (x + 1)(x^2 - x + 4)

Step 2: Partial fraction decomposition:

1 / [(x + 1)(x^2 - x + 4)] = A / (x + 1) + (Bx + C) / (x^2 - x + 4)

Multiply both sides by denominator:

1 = A(x^2 - x + 4) + (Bx + C)(x + 1)

Expand:
1 = A x^2 - A x + 4 A + B x^2 + B x + C x + C

Group terms:
1 = (A + B) x^2 + (-A + B + C) x + (4 A + C)

Equate coefficients:

Coefficient of x^2: 0 = A + B
Coefficient of x: 0 = -A + B + C
Constant term: 1 = 4 A + C

From first: B = -A
From second: 0 = -A + (-A) + C => 0 = -2A + C => C = 2A
From third: 1 = 4 A + C = 4 A + 2 A = 6 A => A = 1/6
Then B = -1/6, C = 2*(1/6) = 1/3

Step 3: Write integral as:

∫ [1/(x + 1)] * (1/6) dx + ∫ [( -1/6 x + 1/3 ) / (x^2 - x + 4)] dx

= (1/6) ∫ 1/(x + 1) dx + ∫ [(-1/6 x + 1/3) / (x^2 - x + 4)] dx

Step 4: Integrate first term:
(1/6) ln|x + 1|

Step 5: For second integral, complete the square in denominator:

x^2 - x + 4 = (x - 1/2)^2 + (4 - 1/4) = (x - 1/2)^2 + 15/4

Let u = x - 1/2

Rewrite numerator:
-1/6 x + 1/3 = -1/6 (u + 1/2) + 1/3 = -1/6 u - 1/12 + 1/3 = -1/6 u + 1/4

So integral becomes:
∫ [(-1/6 u + 1/4) / (u^2 + 15/4)] du

Split integral:
-1/6 ∫ u / (u^2 + 15/4) du + 1/4 ∫ 1 / (u^2 + 15/4) du

First integral:
Let w = u^2 + 15/4, dw = 2u du => u du = dw/2

So,
∫ u / (u^2 + 15/4) du = (1/2) ∫ dw / w = (1/2) ln|w| + C = (1/2) ln(u^2 + 15/4) + C

Therefore,
-1/6 ∫ u / (u^2 + 15/4) du = -1/6 * (1/2) ln(u^2 + 15/4) = -1/12 ln(u^2 + 15/4)

Second integral:
∫ 1 / (u^2 + a^2) du = (1/a) tan^{-1}(u/a) + C

Here, a^2 = 15/4 => a = sqrt(15)/2

So,
1/4 ∫ 1 / (u^2 + (sqrt(15)/2)^2) du = (1/4) * (2 / sqrt(15)) tan^{-1} (2 u / sqrt(15)) = (1 / (2 sqrt(15))) tan^{-1} (2 u / sqrt(15))

Step 6: Substitute back u = x - 1/2

Final answer:

∫ dx / (x^3 + 3x + 4) = (1/6) ln|x + 1| - (1/12) ln[(x - 1/2)^2 + 15/4] + (1 / (2 sqrt(15))) tan^{-1} \left( \frac{2(x - 1/2)}{\sqrt{15}} \right) + C — Factor the cubic denominator, perform partial fraction decomposition, complete the square in the quadratic term, and integrate using standard formulas for logarithmic and inverse tangent integrals.

Question 8

∫ \frac{dx}{x^3 - x^2 + x - 1}

Accepted Answer

First, factorize the denominator:

x^3 - x^2 + x - 1 = (x^3 - x^2) + (x - 1) = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)

So,
∫ dx / (x^3 - x^2 + x - 1) = ∫ dx / [(x - 1)(x^2 + 1)]

Use partial fractions:

1 / [(x - 1)(x^2 + 1)] = A / (x - 1) + (B x + C) / (x^2 + 1)

Multiply both sides by denominator:

1 = A(x^2 + 1) + (B x + C)(x - 1)

Expand:
1 = A x^2 + A + B x^2 - B x + C x - C

Group terms:
1 = (A + B) x^2 + (-B + C) x + (A - C)

Equate coefficients:

x^2: 0 = A + B
x: 0 = -B + C
constant: 1 = A - C

From x^2: B = -A
From x: 0 = -(-A) + C => 0 = A + C => C = -A
From constant: 1 = A - (-A) = A + A = 2A => A = 1/2
Then B = -1/2, C = -1/2

Therefore,

∫ dx / (x^3 - x^2 + x - 1) = ∫ [1/2 / (x - 1)] dx + ∫ [(-1/2 x - 1/2) / (x^2 + 1)] dx

= (1/2) ∫ 1/(x - 1) dx - (1/2) ∫ (x + 1) / (x^2 + 1) dx

Integrate first term:
(1/2) ln|x - 1|

Second integral:
∫ (x + 1) / (x^2 + 1) dx = ∫ x / (x^2 + 1) dx + ∫ 1 / (x^2 + 1) dx

First part:
Let u = x^2 + 1, du = 2x dx => x dx = du/2

So,
∫ x / (x^2 + 1) dx = (1/2) ∫ du / u = (1/2) ln|x^2 + 1|

Second part:
∫ 1 / (x^2 + 1) dx = tan^{-1} x

Therefore,
∫ (x + 1) / (x^2 + 1) dx = (1/2) ln(x^2 + 1) + tan^{-1} x + C

Putting all together:

∫ dx / (x^3 - x^2 + x - 1) = (1/2) ln|x - 1| - (1/2) [(1/2) ln(x^2 + 1) + tan^{-1} x] + C

= (1/2) ln|x - 1| - (1/4) ln(x^2 + 1) - (1/2) tan^{-1} x + C — Factor the denominator, apply partial fractions, integrate logarithmic and inverse tangent terms separately.

Question 9

∫ \frac{dx}{x - 1}

Accepted Answer

Integral of 1/(x - 1) dx is:

∫ 1/(x - 1) dx = ln|x - 1| + C — Standard integral of 1/(x - a) is ln|x - a| + C.

Question 10

∫ (1 - x)/x dx

Accepted Answer

Rewrite the integrand:

(1 - x)/x = 1/x - x/x = 1/x - 1

So,
∫ (1 - x)/x dx = ∫ (1/x - 1) dx = ∫ 1/x dx - ∫ 1 dx

Integrate each term:
∫ 1/x dx = ln|x|
∫ 1 dx = x

Therefore,
∫ (1 - x)/x dx = ln|x| - x + C — Split the fraction and integrate term by term.

Question 11

∫ x (3 x^2 + 2 x + 3) dx

Accepted Answer

First, expand the integrand:

x (3 x^2 + 2 x + 3) = 3 x^3 + 2 x^2 + 3 x

Now integrate term by term:

∫ 3 x^3 dx = 3 (x^4 / 4) = (3/4) x^4
∫ 2 x^2 dx = 2 (x^3 / 3) = (2/3) x^3
∫ 3 x dx = 3 (x^2 / 2) = (3/2) x^2

Therefore,

∫ x (3 x^2 + 2 x + 3) dx = (3/4) x^4 + (2/3) x^3 + (3/2) x^2 + C — Multiply out the terms and integrate each using power rule.

Question 12

∫ (2 x - 3 cos x + e^x) dx

Accepted Answer

Integrate each term separately:

∫ 2 x dx = 2 (x^2 / 2) = x^2
∫ -3 cos x dx = -3 ∫ cos x dx = -3 sin x
∫ e^x dx = e^x

Therefore,
∫ (2 x - 3 cos x + e^x) dx = x^2 - 3 sin x + e^x + C — Use standard integrals for polynomial, trigonometric, and exponential functions.

Question 13

∫ (2 x^2 - 3 sin x + 5 x) dx

Accepted Answer

Integrate each term separately:

∫ 2 x^2 dx = 2 (x^3 / 3) = (2/3) x^3
∫ -3 sin x dx = -3 (-cos x) = 3 cos x
∫ 5 x dx = 5 (x^2 / 2) = (5/2) x^2

Therefore,
∫ (2 x^2 - 3 sin x + 5 x) dx = (2/3) x^3 + 3 cos x + (5/2) x^2 + C — Use power rule for polynomial terms and standard integrals for sine function.

Question 14

∫ sec x (sec x + tan x) dx

Accepted Answer

Rewrite the integrand:

sec x (sec x + tan x) = sec^2 x + sec x tan x

We know that:

d/dx (tan x) = sec^2 x
and
d/dx (sec x) = sec x tan x

Therefore,

∫ sec x (sec x + tan x) dx = ∫ (sec^2 x + sec x tan x) dx = ∫ d/dx (tan x) dx + ∫ d/dx (sec x) dx = tan x + sec x + C — Recognize the integrand as sum of derivatives of tan x and sec x and integrate accordingly.

Question 15

∫ sec^2 x dx

Accepted Answer

Integral of sec^2 x is:

∫ sec^2 x dx = tan x + C — Standard integral of sec^2 x is tan x + C.

Integrals

Integrals — Study Notes

Introduction

Integral as an Antiderivative

Basic Integration Formulas

Practice Questions — Integrals

All 7 Chapters in Mathematics Part-II