AMITY INTERNATIONAL SCHOOL
Power Grid Complex Sec - 43, Gurgaon
Class: XII
Session: 2021-22
Integrals, Application of Integrals and Differential Equations
Indefinite Integrations
Different forms of integrations: Concepts and Formulae
Integration by simple rearrangements.
Use simple trigonometric and algebraic formulas or simple rearrangements to solve it
for example: (i) ∫ tan 𝑥 tan 2𝑥 tan 3𝑥 dx.
• Integration by simple substitutions.
• Integral of the form ∫
𝑓 ′ (𝑥)
𝑑𝑥 = log{𝑓(𝑥)} + 𝐶
𝑓(𝑥)
• Integral of the form ∫{𝑓(𝑥)}𝑛 𝑓′(𝑥)𝑑𝑥
𝑓(𝑥)
• Integral of the form ∫ 𝑓′(𝑥) 𝑑𝑥
• Integral of the form∫ 𝒔𝒊𝒏𝒎 𝒙𝒅𝒙 , ∫ 𝒄𝒐𝒔𝒎 𝒙 dx. Where m≤ 4,use the following trigonometrical
identities: (i) 𝑠𝑖𝑛2 𝑥 =
1−𝑐𝑜𝑠 2𝑥
(ii) 𝑐𝑜𝑠 2 𝑥 =
2
3
(iii) 𝑠𝑖𝑛 3𝑥 = 3𝑠𝑖𝑛 𝑥 − 4𝑠𝑖𝑛 𝑥
1+𝑐𝑜𝑠 2𝑥
2
(iv) 𝑠𝑖𝑛 3𝑥 = 4𝑐𝑜𝑠 3 𝑥 − 3 𝑐𝑜𝑠 𝑥.
• Integral of the form ∫ 𝒔𝒊𝒏 𝒎𝒙 𝒄𝒐𝒔 𝒏𝒙 𝒅𝒙, ∫ 𝒔𝒊𝒏 𝒎𝒙 𝒔𝒊𝒏 𝒏𝒙 𝒅𝒙 , ∫ 𝒄𝒐𝒔 𝒎𝒙 𝒄𝒐𝒔 𝒏𝒙 𝒅𝒙.
To evaluate this type of integrals we use the following trigonometrical identities:
2sin A cos B = sin(A + B) + sin(A − B) ; 2cos A sin B = sin(A + B) − sin(A − B)
2cos A cos B = cos(A + B) + cos(A − B) ; 2sin A sin B = cos(A − B) − cos(A + B)
• Integral of the form ∫ 𝐬𝐢𝐧𝐦 𝐱 𝐜𝐨𝐬𝐧 𝐱 dx, m, n 𝝐 N.
Conditions
Substitutions
i.
If 𝑚 is an odd (+)ive integer and n is even
𝑐𝑜𝑠 𝑥 = 𝑡
ii.
If 𝑛 is an odd (+)ive integer
𝑠𝑖𝑛 𝑥 = 𝑡
iii.
If both 𝑚, 𝑛 are odd (+) ive integers
either 𝑠𝑖𝑛 𝑥 = 𝑡 𝑜𝑟 𝑐𝑜𝑠 𝑥 = 𝑡.
iv.
If both 𝑚, 𝑛 are even (+) ive integers
use multiple angle formula
v.
If 𝑚 + 𝑛 is a negative even integer
𝑡𝑎𝑛 𝑥 = 𝑡
•
∫ sinn x dx, where n is (+)ive odd integer, put cos x = t
•
∫ cosn x dx, where n is (+)ive odd integer, put sin x = t
•
∫ sec n x dx, where n is (+)ive even integer, put tan x = t
•
Integral of the forms
𝐝𝐱
𝐝𝐱
𝐝𝐱
𝐝𝐱
𝐝𝐱
∫ 𝐚 𝐬𝐢𝐧𝟐 𝐱 + 𝐛 𝐜𝐨𝐬 𝟐 𝐱 , ∫ 𝐚 + 𝐛 𝐬𝐢𝐧𝟐 𝐱 , ∫ 𝐚 + 𝐛 𝐜𝐨𝐬𝟐 𝐱 , ∫ ( 𝐚 𝐬𝐢𝐧𝐱 + 𝐛 𝐜𝐨𝐬𝐱 )𝟐 , ∫ 𝐚+ 𝐛 𝐬𝐢𝐧𝟐 𝐱 + 𝐜 𝐜𝐨𝐬𝟐 𝐱
(i) Divide the numerator and denominator both by cos 2 x..
(ii) Replace sec2x, if any, in denominator by 1 + tan2 x.
(iii) Put tan x = t so that sec 2 x dx = dt.
dt
This substitution reduces the integral in the form ∫ at2+bt+c. Now evaluate the integral obtained.
•
Integral of the forms:
𝐝𝐱
𝐝𝐱
𝐝𝐱
𝐝𝐱
∫ 𝐚 𝐬𝐢𝐧𝐱 + 𝐛 𝐜𝐨𝐬𝐱 , ∫ 𝐚 + 𝐛 𝐬𝐢𝐧𝐱 , ∫ 𝐚 + 𝐛 𝐜𝐨𝐬𝐱 , ∫ 𝐚 + 𝐛 𝐬𝐢𝐧𝐱 + 𝐜 𝐜𝐨𝐬𝐱
1−tan2 x⁄2
(i) Put cos x =
and sin x =
1+tan2 x⁄2
2 tanx⁄2
1+tan2 x⁄2
(ii) Replace (1+tan2 x⁄2) in the numerator by sec 2 x⁄2
x
(iii) Put tan 2 = t so that
1
2
sec 2 x⁄2 dx = dt. , by this substitution the integral reduces in
dt
the form ∫ at2+bt+c. Now evaluate the integral obtained.
•
Integral of the forms:
dx
dx
dx
dx
∫ ax2+ bx + c ,∫ √ax2+ bx + c , ∫ c + bx − ax2 , ∫ √c + bx−ax2
Make the coefficient of x2 unity, if it is not, then make the denominator as a perfect square and reduce it
in the form [ (x±𝛼)2 ± 𝛽2 ], and apply the following formulas:
1
1
𝑥
1
1
𝑎+𝑥
1
1
𝑥−𝑎
∫ 𝑎2 +𝑥 2 𝑑𝑥 = 𝑎 tan−1 ( 𝑎 ) +c , ∫ 𝑎2 −𝑥 2 𝑑𝑥 = 2𝑎 log |𝑎−𝑥| + c , ∫ 𝑥 2 −𝑎2 𝑑𝑥 = 2𝑎 log |𝑥+𝑎| +c
1
1
∫ √𝑎2+𝑥 2 𝑑𝑥 = log|𝑥 + √𝑎2 + 𝑥 2 | +c , ∫ √𝑥 2−𝑎2 𝑑𝑥 = log|𝑥 + √𝑥 2 − 𝑎2 | + c ,
1
𝑥
∫ √𝑎2−𝑥 2 𝑑𝑥 = sin−1 ( 𝑎 ) + c
•
Integral of the forms:
∫(px + q)√ax 2 + bx + c dx, ∫
(i) Write (px + q) as:
px+q
ax2 + bx + c
dx, ∫
px+q
√ax2
+ bx +c
dx.
𝑑
px + q = A 𝑑𝑥(ax2+bx+c) +B = A (2ax + b) + B.
(ii) Obtain the values of A and B by equating the coefficients of x and constant terms.
(iii) Replace ( px + q ) by A (2ax + b) + B in the given integrals to get ,
𝑝𝑥+𝑞
2𝑎𝑥+𝑏
1
∫ 𝑎𝑥 2+ 𝑏𝑥+𝑐 dx= A ∫ 𝑎𝑥 2+ 𝑏𝑥+𝑐dx + B ∫ 𝑎𝑥 2+ 𝑏𝑥+𝑐 dx.
px+q
2ax+b
and
1
∫ √ax2 + bx +cdx = A ∫ √ax2 + bx +c dx + B ∫ √ax2 + bx +c dx.
•
∫(px + q)√ax 2 + bx + c dx = A ∫(2𝑎𝑥 + 𝑏) √ax 2 + bx + c dx + B ∫ √ax 2 + bx + c dx . Now integrate
Integral of the forms:
x2 +A
x2 −A
1
∫ x4+ kx2 + A2 dx, ∫ x4 + kx2 + A2 dx, ∫ x4 +kx2+1 dx, ∫ √tan x dx, ∫ √cot 𝑥 dx. A >0, k 𝜖 𝑅
(i) Divide the numerator and denominator by x2.
A
𝐴
A
Introduce (1 + x2 ) or (1 − 𝑥 2 ) or both in the numerator. And the denominator in the form of ( x ± x )2 ∓ α2
𝐴
A
𝐴
𝐴
Substitute (x + 𝑥 ) = t when numerator is (1− x2 ).
Substitute (x − 𝑥 ) = t when numerator is (1+ 𝑥 2 ).
The integral reduces in one of the following forms
1
1
∫ 𝑥 2 + 𝑎2 dx , ∫ 𝑥 2 + 𝑎2 dx. Now integrate.
•
Integrating by parts
Integral of the product of two functions = First function x Integral of second – Integral of (derivative of
first x integral of second).
du
∫ uv dx = u ∫ v dx – ∫( dx ∫ v dx ) dx .
If two functions are of different types, take the first function to be the function which comes first in the
word ‘ILATE ′ where
i.
ii.
iii.
iv.
v.
I
L
A
T
E
for inverse trigonometric function.
for the logarithmic function.
for algebraic function.
for trigonometric function.
for exponential function.
𝐟(𝐱)
• Partial Fraction 𝐠(𝐱) form:
If degree of f(x) < degree of g(x) then it is a proper fraction.
If degree of f(x) ≥ degree of g(x) then it is an improper fraction, divide the numerator by denominator and
then simplify and integrate.
S. No.
Form of the rational function
𝑝𝑥 + 𝑞
(𝑥−𝑎)(𝑥−𝑏)
1.
𝑝𝑥 + 𝑞
(𝑥 − 𝑎)2
2.
3.
,𝑎 ≠ 𝑏
𝑝𝑥 2 + 𝑞𝑥 + 𝑟
,𝑎 ≠ 𝑏 ≠ 𝑐
(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)
4.
𝑝𝑥 2 + 𝑞𝑥 + 𝑟
(𝑥 − 𝑏)(𝑥 − 𝑎)2
5.
𝑝𝑥 2 + 𝑞𝑥 + 𝑟
(𝑥 − 𝑎)(𝑥 2 + 𝑏𝑥 + 𝑐)
Form of the partial fraction
𝐴
(𝑥−𝑎)
+ (𝑥−𝑏)
𝐴
(𝑥−𝑎)
+ (𝑥−𝑎)2
𝐴
(𝑥−𝑎)
𝐴
(𝑥−𝑎)
𝐵
𝐵
𝐵
𝐶
𝐵
𝐶
+ (𝑥−𝑏) + (𝑥−𝑐)
+ (𝑥−𝑎)2 + (𝑥−𝑏)
𝐴
(𝑥−𝑎)
𝐵𝑥 + 𝐶
+ 𝑥 2 +𝑏𝑥+𝑐
Where 𝒙𝟐 + 𝒃𝒙 + 𝒄 cannot
be factorized further
•
Some Special Integrals
∫
sin 𝑥+cos 𝑥
√sin 2𝑥
dx
, ∫
sin 𝑥− cos 𝑥
√sin 2𝑥
dx
Put the additive inverse of the numerator as t, and simplify the denominator by making the square of the
substitution. As in the 1st example,
put (𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥) = 𝑡 so that (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥) 𝑑𝑥 = 𝑑𝑡
And (𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥)2 = 𝑡 2
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 − 2𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 = 𝑡 2 or
2
1 − 𝑠𝑖𝑛2𝑥 = 𝑡 . Now substitute the values and integrate.
Some Important Indefinite Integration Formulae:
1. ∫ 𝑥 𝑛 𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
+ C, 𝑛 ≠ −1
2. ∫ 𝑑𝑥 = 𝑥 + 𝐶
3. ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
4. ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶
5. ∫ 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
6. ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑥𝑑𝑥 = − cot 𝑥 + 𝐶
7. ∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
8. ∫ cosec 𝑥 cot 𝑥 𝑑𝑥 = − cosec 𝑥 + 𝐶
1
9. ∫
𝑑𝑥 = sin−1 𝑥+ C
√1− 𝑥 2
1
10. ∫
𝑑𝑥 = −cos−1 𝑥+ C
√1− 𝑥 2
𝑑𝑥
11. ∫ 1+𝑥2 = tan−1 𝑥 + C
𝑑𝑥
12. ∫ 1+𝑥2 = − cot −1 𝑥 + C
13. ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
1
14. ∫ 𝑥 𝑑𝑥 = log|𝑥| + 𝐶
15. ∫ 𝑎 𝑥 𝑑𝑥 =
𝑎𝑥
log 𝑎
+C
1
16. ∫ 𝑑𝑥 = log 𝑥 + 𝐶
𝑥
17. ∫
1
𝑑𝑥
√𝑥
= 2√𝑥 + 𝐶
18. ∫ 𝑡𝑎𝑛 𝑥 𝑑𝑥 = 𝑙𝑜𝑔 |𝑠𝑒𝑐𝑥| + 𝐶 = − 𝑙𝑜𝑔 |𝑐𝑜𝑠𝑥| + 𝐶
19. ∫ 𝑐𝑜𝑡 𝑥 𝑑𝑥 = 𝑙𝑜𝑔 |𝑠𝑖𝑛𝑥| + 𝐶
20. ∫ 𝑠𝑒𝑐 𝑑𝑥 = 𝑙𝑜𝑔 |𝑠𝑒𝑐 𝑥 + 𝑡𝑎𝑛𝑥| + 𝐶
21. ∫ 𝑐𝑜𝑠𝑒𝑐 𝑑𝑥 = 𝑙𝑜𝑔 |𝑐𝑜𝑠𝑒𝑐𝑥 − 𝑐𝑜𝑡𝑥| +𝐶
1
1
𝑥
1
1
𝑎+𝑥
22. ∫ 𝑎2 +𝑥 2 𝑑𝑥 = 𝑎 tan−1 ( 𝑎 ) +C ,
23. ∫ 2 2 𝑑𝑥 = log | | + C ,
𝑎 −𝑥
2𝑎
𝑎−𝑥
1
1
𝑥−𝑎
24. ∫ 𝑥 2 −𝑎2 𝑑𝑥 = 2𝑎 log |𝑥+𝑎| +C ,
1
25. ∫ 2 2 𝑑𝑥 = log|𝑥 + √𝑎2 + 𝑥 2 | +C ,
√𝑎 +𝑥
1
26. ∫ 2 2 𝑑𝑥 = log|𝑥 + √𝑥 2 − 𝑎2 | + C ,
√𝑥 −𝑎
1
𝑥
27. ∫ 2 2 𝑑𝑥 = sin−1 ( 𝑎 ) + C.
√𝑎 −𝑥
𝑥
𝑎2
log|𝑥
2
+ √𝑥 2 − 𝑎2 | + 𝐶
𝑥
𝑎2
log|𝑥
2
+ √𝑥 2 + 𝑎2 | + 𝐶
𝑥
𝑎2
𝑥
sin−1 (𝑎) + C
2
28. ∫ √𝑥 2 − 𝑎2 𝑑𝑥 = 2 √𝑥 2 − 𝑎2 −
29. ∫ √𝑥 2 + 𝑎2 𝑑𝑥 = 2 √𝑥 2 + 𝑎2 +
30. ∫ √𝑎2 − 𝑥 2 𝑑𝑥 = 2 √𝑎2 − 𝑥 2 +
Definite integration
𝑏
➢ Fundamental Theorem of Calculus: We have defined ∫𝑎 𝑓(𝑥)𝑑𝑥 as the area of the region bounded by
the curve 𝑦 = 𝑓(𝑥), the ordinates 𝑥 = 𝑎 and 𝑥 = 𝑏 and 𝑥-axis. Let 𝑥 be a given point in [𝑎, 𝑏]. Then
𝑥
∫𝑎 𝑓(𝑥)𝑑𝑥 represents the area of the light shaded region
➢ First fundamental theorem of integral calculus:
Theorem - 1
Let 𝑓 be a continuous function on the closed interval [𝑎, 𝑏] and let 𝐴 (𝑥) be the area
function. Then 𝐴′(𝑥) = ∫ 𝑓 (𝑥), for all 𝑥 ∈ [𝑎, 𝑏].
➢ Second fundamental theorem of integral calculus:
Theorem - 2
Let f be continuous function defined on the closed interval [𝑎, 𝑏] and 𝐹 be an anti
𝑏
derivative of 𝑓. Then ∫𝑎 𝑓(𝑥)𝑑𝑥dx = [𝐹(𝑥)]𝑏𝑎 = F (b) – F(a).
PROPERTIES OF DEFINITE INTEGRALS:
Application Of Integrals (Graphs of Some Standard Curves)
Parabolas
Circles
Ellipse
Modulus
Functions
𝒚 = 𝒙𝟑
Formula Used
𝑥 𝑛+1
𝑏
𝑏
➢ ∫𝑎 𝑥 𝑛 𝑑𝑥 = [
]
𝑛+1
𝑎
𝑑
𝑥
➢ ∫𝑐 √𝑎2 − 𝑥 2 𝑑𝑥 = [2 √𝑎2 − 𝑥 2 +
𝑎2
2
𝑥
sin−1 (𝑎)]
𝑑
𝑐
➢ Equation of Straight Line passing through two points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 )is
𝑦 − 𝑦1 = (
𝑦2 − 𝑦1
) (𝑥 − 𝑥1 )
𝑥2 − 𝑥1
➢ Area under Simple Curves
➢ The area of the region bounded by the curve 𝑦 = 𝑓 (𝑥), 𝑥-axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏
(𝑏 > 𝑎) is given by the formula:
𝒃
𝒃
Area = ∫𝒂 𝒚𝒅𝒙 = ∫𝒂 𝒇(𝒙)𝒅𝒙
➢ The area of the region bounded by the curve 𝑥 = 𝑔(𝑦), 𝑦-axis and the lines
𝑦 = 𝑐, 𝑦 = 𝑑 is given by the formula:
𝒅
𝒅
Area = ∫𝒄 𝒙𝒅𝒚 = ∫𝒄 𝒈(𝒚)𝒅𝒚
➢ The area of the region enclosed between two curves 𝑦 = 𝑓 (𝑥), 𝑦 = 𝑔 (𝑥) and the
lines 𝑥 = 𝑎, 𝑥 = 𝑏 is given by the formula.
𝒃
Area = ∫𝒂 (𝒇(𝒙) − 𝒈(𝒙))𝒅𝒙
KEY POINTS :
𝑑
𝑥
𝑎2
𝑥
𝑑
•
∫𝑐 √𝑎2 − 𝑥 2 𝑑𝑥 = [2 √𝑎2 − 𝑥 2 +
•
•
•
•
Curve 𝑎𝑥 + 𝑏𝑦 = 𝑐 is a straight line
Curve 𝑦 2 = 4𝑎𝑥 is a parabola with vertex (0,0) and axis is x-axis.
Curve 𝑥 2 = 4𝑎𝑦 is a parabola with vertex (0,0) and axis is y-axis.
Curve 𝑥 2 +𝑦 2 = 𝑎2 is a circle with center (0,0) and radius = 𝑎
•
Curve 𝑎2 + 𝑏2 = 1, 𝑎 > 𝑏 and 𝑎2 + 𝑏2 = 1, 𝑎 > 𝑏 is an ellipse with centre (0,0)
•
𝑥2
𝑦2
𝑦2
2
sin−1 (𝑎)]
𝑐
𝑥2
Circle and ellipse are symmetric in each quadrant.
DIFFERENTIAL EQUATIONS
𝑑𝑦
➢ For a given function 𝑔, a function 𝑓 such that 𝑑𝑥 = 𝑔(𝑥), where 𝑦 = 𝑓(𝑥)
An equation of this form is known as a differential equation
➢ Order of a differential equation: is defined as the order of the highest order derivative of the
dependent variable with respect to the independent variable involved in the given differential equation.
➢ Degree of a differential equation: is the degree of the highest order derivative, when differential
coefficients are made free from radicals and fractions
➢ NOTE: If differential equation involves following types of terms, then we say that degree of
differential equation is not defined- 𝐬𝐢𝐧 𝒚′ , 𝐜𝐨𝐬 𝒚′, 𝑒 𝑦′ , 𝒍𝒐𝒈 𝒚′ 𝒆𝒕𝒄.
➢ Solution of Differential Equations• The solution which contains arbitrary constants is called the general solution (primitive) of the differential
equation.
• The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving
particular values to the arbitrary constants is called a particular solution of the differential equation.
➢ NOTE: The order of a differential equation representing a family of curves is same as the number of
arbitrary constants present in the equation corresponding to the family of curves
➢ Methods of Solving First Order, First Degree Differential Equations
➢ Equations in variable separable form: Variable separable method’ is used to solve such equation in
which variables can be separated completely, i.e. terms containing 𝑥 should remain with 𝑑𝑥 and terms
containing 𝑦 should remain with 𝑑𝑦.
➢ Homogeneous Differential Equation:
A function 𝐹 (𝑥, 𝑦) is a homogeneous function of degree 𝑛 if 𝐹 (𝜆𝑥, 𝜆𝑦 ) = 𝜆𝑛 𝐹 (𝑥, 𝑦)
for some non-zero constant 𝜆. A differential equation which can be expressed in the form
𝑑𝑦
𝑑𝑥
= 𝐹 (𝑥, 𝑦) or 𝑑𝑦 = 𝐺 (𝑥, 𝑦), where 𝐹 (𝑥, 𝑦) and 𝐺 (𝑥, 𝑦) are homogeneous functions of
𝑑𝑥
degree zero,
➢ Steps to solve a homogeneous differential equation of the type
𝒅𝒚
𝒚
(i) 𝒅𝒙 = 𝑭(𝒙, 𝒚) = 𝒈 (𝒙)... (1)
(ii) Put 𝒚 = 𝒗 𝒙 ... (2)
𝒅𝒚
𝒅𝒗
(iii) Differentiating equation (2) with respect to 𝑥, we get 𝒅𝒙 = 𝒗 + 𝒙 𝒅𝒙... (3)
(iv) Put the value of
𝒅𝒗
𝑑𝑦
𝑑𝑥
from equation (3) in equation (1), we get
𝒅𝒗
𝒗 + 𝒙 𝒅𝒙 = 𝒈 (𝒗) 𝑜𝑟 𝒙 𝒅𝒙 = 𝒈 (𝒗) – 𝒗 ... (4)
𝑦
Now apply Variable separable method to integrate and replace 𝑣 by 𝑥
•
𝑑𝑥
If the equation is of the form 𝑑𝑦 = 𝐺 (𝑥, 𝑦) We make the substitution 𝑥 = 𝑣 𝑦... (2)
𝑑𝑥
𝑑𝑣
Differentiating equation (2) with respect to 𝑦, we get 𝑑𝑦 = 𝑣 + 𝑦 𝑑𝑦...(3)proceed in the same way.
𝑑𝑦
➢ Linear differential equations: Is of the from 𝑑𝑥 + 𝑃𝑦 = 𝑄 where, P and Q are constants
or functions of 𝑥 only, is known as a first order linear differential equation.
➢ Steps to solve first order linear differential equation
(i) Express the given differential equation in the form
𝒅𝒚
𝒅𝒙
(ii) Find the Integrating Factor (I.F) = 𝒆∫ 𝑷𝒅𝒙 .
(iii) Write the solution of the given differential equation as
+ 𝑷𝒚 = 𝑸
y (I.F) = ∫ 𝑸 × 𝑰. 𝑭 𝒅𝒙 + 𝑪
➢ Note:- Before solving the questions, note the following concepts: (here the base of log is e.)
1
(𝑎) 𝑒log 𝑥 = 𝑥
(𝑏) 𝑒− log 𝑥 = 𝑒log 𝑥−1= 𝑥−1 =𝑥
(c) 𝑒2 log 𝑥 = 𝑒log 𝑥2 = 𝑥2
(𝑑) 𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑒log 𝑓(𝑥) = 𝑓(𝑥)