Integration questions are an essential part of calculus and are widely used in Class 11, 12, and advanced mathematics. Just like differentiation helps us find rates and slopes, integration is the reverse process, helping us calculate areas, volumes, and accumulated quantities.
Whether you're preparing for CBSE, ICSE, JEE, NEET, or any competitive exam, understanding how to do integration and practicing integrals questions is crucial.
Table of Contents
What is integration? In simple words, integration is the process of adding up small parts to find the whole. In mathematics, it refers to summing up infinitesimally small quantities to find areas under curves, total distance, or volume.
It is the inverse of differentiation. If
d/dx (F(x)) = f(x)
Then
∫ f(x) dx = F(x) + C
Where C is the constant of integration.
You must understand what is integration to solve problems in physics, mathematics, and engineering efficiently.
Let’s define integration properly:
Integration is a fundamental operation in calculus that combines infinitesimal data into a whole quantity. It helps determine the total accumulation of a quantity.
Example:
∫ x² dx = (x³)/3 + C
So whenever someone asks to define integration, say it's the process of finding an integral of a function which results in accumulation or area.
Memorizing integral formulas is key to solving integration questions quickly.
Common Integral Formulas Table
Function |
Integral |
∫ xⁿ dx (n ≠ -1) |
xⁿ⁺¹ / (n + 1) + C |
∫ 1/x dx |
ln |
∫ eˣ dx |
eˣ + C |
∫ aˣ dx |
aˣ / ln(a) + C |
∫ sin(x) dx |
-cos(x) + C |
∫ cos(x) dx |
sin(x) + C |
∫ sec²(x) dx |
tan(x) + C |
∫ csc²(x) dx |
-cot(x) + C |
∫ sec(x)·tan(x) dx |
sec(x) + C |
∫ csc(x)·cot(x) dx |
-csc(x) + C |
∫ tan(x) dx |
ln |
∫ cot(x) dx |
ln |
∫ 1 / √(1 − x²) dx |
sin⁻¹(x) + C |
∫ 1 / √(1 + x²) dx |
sinh⁻¹(x) + C or ln |
∫ 1 / (x² + a²) dx |
(1/a)·tan⁻¹(x/a) + C |
∫ 1 / (a² − x²) dx |
(1/2a)·ln |
Revise these integral formulas regularly to ace your integration questions.
Question:
Solve ∫ 2x(x² + 1)³ dx
Step 1: Let u = x² + 1
Then, du/dx = 2x → du = 2x dx
Step 2: Replace in the integral:
∫ 2x(x² + 1)³ dx = ∫ u³ du
Step 3: Integrate:
∫ u³ du = u⁴ / 4 + C
Step 4: Substitute back u = x² + 1
Final Answer: (x² + 1)⁴ / 4 + C
Question:
Solve ∫ x·e^x dx
Step 1: Let
u = x → du = dx
dv = e^x dx → v = e^x
Step 2: Use integration by parts formula:
∫ u dv = uv − ∫ v du
Apply the formula:
= x·e^x − ∫ e^x dx
= x·e^x − e^x + C
Final Answer:
x·e^x − e^x + C
Question:
Solve ∫ 1/[(x + 1)(x + 2)] dx
Step 1: Express the integrand in partial fractions:
1/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)
Multiply both sides by (x + 1)(x + 2):
1 = A(x + 2) + B(x + 1)
Now solve for A and B:
Let x = −2 → 1 = A(0) + B(−1) → B = −1
Let x = −1 → 1 = A(1) + B(0) → A = 1
So:
∫ 1/[(x + 1)(x + 2)] dx = ∫ [1/(x + 1) − 1/(x + 2)] dx
Step 2: Integrate each term:
= ln|x + 1| − ln|x + 2| + C
= ln|(x + 1)/(x + 2)| + C
Final Answer:
ln|(x + 1)/(x + 2)| + C
Question:
Solve ∫ sin²x dx
Step 1: Use the identity:
sin²x = (1 − cos(2x)) / 2
Step 2: Rewrite the integral:
∫ sin²x dx = ∫ (1 − cos(2x))/2 dx
= (1/2) ∫ (1 − cos(2x)) dx
Step 3: Integrate:
= (1/2)[x − (1/2)sin(2x)] + C
= x/2 − sin(2x)/4 + C
Final Answer:
x/2 − sin(2x)/4 + C
Step 1: Use the power rule of integration:
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C
Step 2: Substitute n = 2
∫ x² dx = x³ / 3 + C
Final Answer:
x³ / 3 + C
Step 1: Use the standard formula:
∫ 1/x dx = ln|x| + C
Final Answer:
ln|x| + C
Step 1: Apply the exponential rule:
∫ e^x dx = e^x + C
Final Answer:
e^x + C
Step 1: Use the trigonometric formula:
∫ cos(x) dx = sin(x) + C
Final Answer:
sin(x) + C
Step 1: Use integration by parts. Let:
u = x
dv = cos(x) dx
Then:
du = dx
v = sin(x)
Step 2: Apply the formula:
∫ u dv = u·v − ∫ v du
= x·sin(x) − ∫ sin(x) dx
= x·sin(x) + cos(x) + C
Final Answer:
x·sin(x) + cos(x) + C
Step 1: Use the standard formula:
∫ 1 / (x² + 1) dx = tan⁻¹(x) + C
Final Answer:
tan⁻¹(x) + C
Step 1: Integrate sin(x):
∫ sin(x) dx = −cos(x)
Step 2: Apply the limits from 0 to π:
= [−cos(x)]₀^π
= −cos(π) + cos(0)
= −(−1) + 1 = 2
Final Answer: 2
Step 1: Use integration by parts. Let:
u = ln(x)
dv = dx
Then:
du = 1/x dx
v = x
Step 2: Apply the formula:
∫ u dv = u·v − ∫ v du
= x·ln(x) − ∫ x·(1/x) dx
= x·ln(x) − ∫ 1 dx
= x·ln(x) − x + C
Final Answer:
x·ln(x) − x + C
Formula:
Work = ∫ₐᵇ F(x) dx
Solution:
= ∫₀⁵ 4x dx
= 4 ∫₀⁵ x dx
= 4 [x² / 2]₀⁵
= 4 [(25 / 2) − 0] = 50
Final Answer:
Work = 50 units
Step 1: Use substitution. Let:
u = x² + 1
Ten:
du = 2x dx → x dx = (1/2) du
Step 2: Rewrite the integral:
∫ x·√(x² + 1) dx = (1/2) ∫ √u du
= (1/2) ∫ u^(1/2) du
Step 3: Integrate:
= (1/2) · (2/3) u^(3/2)
= (1/3) u^(3/2)
Step 4: Substitute back u = x² + 1
= (1/3) · (x² + 1)^(3/2) + C
Final Answer:
(1/3)(x² + 1)^(3/2) + C
Learn and revise all integral formulas
Identify if substitution, parts, or partial fraction is needed
Always simplify the function before solving
Don’t forget the constant C in indefinite integrals
Practice 10+ integration questions daily
Use a formula sheet for quick reference
∫ sin²x dx
∫ (3x² + 2x + 5) dx
∫ 1/(x² + 1) dx
∫ x·√(x² + 1) dx
∫ tan x dx
∫ ln x dx
∫ (e^x)/(1 + e^x) dx
∫ 1/√(1 - x²) dx
∫ sec x dx
∫ 1/(x² - a²) dx
A car's velocity is given by v(t) = 2t. Find the distance travelled from t = 0 to t = 5.
Water is flowing into a tank at a rate of R(t) = 3t litres per hour. How much water flows into the tank from t = 0 to t = 4?
To conclude, integration questions are a core concept in calculus. By understanding what is integration, being able to define integration, and mastering all the integral formulas, you’ll become efficient in solving all types of integrals questions.
Practice different question types daily and apply what you've learned about how to do integration. The more you practice, the better your speed and confidence will become.
Related Links
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Differentiation Questions - Practice a wide range of differentiation problems, from finding rates of change to solving advanced calculus questions, to enhance your problem‑solving skills.
Here are the 5 basic integration formulas:
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, where n ≠ -1
∫ 1/x dx = ln|x| + C
∫ e^x dx = e^x + C
∫ a^x dx = a^x / ln(a) + C
∫ sin(x) dx = -cos(x) + C
The integral of cos(x) is:
∫ cos(x) dx = sin(x) + C
Apply the power rule:
∫ 2x dx = x² + C
Again, use the power rule:
∫ 3x dx = (3x²) / 2 + C
Use substitution:
∫ e^(2x) dx = (1/2)·e^(2x) + C
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