Integration by Parts MCQ Quiz - Objective Question with Answer for Integration by Parts - Download Free PDF

Concept Used:

Integration by parts:

Calculation:

⇒ Let , then

The integral becomes   

⇒

⇒

Hence option 2 is correct

Consider, ,

Solving the indefinite integral,

,

,

Applying integral by parts,

,

,

,

The value of the indefinite integral,

,

,

,

Calculation

Using integration by parts

Hence option 2 is correct

Concept:

Integration by Substitution:

• When an integral involves a composite function, substitution can simplify it by letting the inner function be a new variable.
• If we have an integral of the form ∫ e^f(x) f'(x) dx, it directly integrates to e^f(x) + C.
• Exponential Function e^x: It is a function equal to its own derivative.
• Important Formula: ∫ e^f(x) f'(x) dx = e^f(x) + C

Calculation:

Given,

Integral =   

Let’s define u = √x

⇒ u = x^1/2

⇒ Differentiating, du/dx = (1/2)x^-1/2

⇒ du = (1/2√x) dx

⇒ dx = 2√x du

Now substitute in the integral,

Integral = ∫ e^x × (2x + 1)/(2√x) × (2√x) du

⇒ Integral = ∫ e^x (2x + 1) du

Now, notice that we still have x terms in e^x and (2x + 1).

Thus, no substitution is needed here actually.

Rewrite the original integral:

Integral = ∫ e^x × (2x + 1)/(2√x) dx

Split the fraction:

Integral = (1/2) ∫ (2x/√x + 1/√x) e^x dx

⇒ (1/2) ∫ (2x^1/2 + x^-1/2) e^x dx

Now let’s define, t = √x

⇒ t = x^1/2

⇒ x = t²

⇒ dx = 2t dt

Substituting all in terms of t,

Integral = (1/2) ∫ (2t + 1/t) e^t² × 2t dt

Expand:

Integral = (1/2) × 2 ∫ (2t² + 1) e^t² dt

⇒ Integral = ∫ (2t² + 1) e^t² dt

Now split the integral:

Integral = ∫ 2t² e^t² dt + ∫ e^t² dt

For ∫ 2t² e^t² dt:

d/dt (e^t²) = 2t e^t²

We need t² terms. So consider:

Let’s differentiate (t e^t²):

⇒ d/dt (t e^t²) = e^t² + t × 2t e^t²

⇒ d/dt (t e^t²) = e^t² + 2t² e^t²

Thus,

e^t² + 2t² e^t² = d/dt (t e^t²)

Thus,

∫ (2t² + 1) e^t² dt = ∫ d/dt (t e^t²) dt

⇒ t e^t² + C

Substituting back t = √x:

⇒ √x e^x + C

∴ Hence, the final answer is  

Concept Used:

Integration by parts:

Calculation:

⇒ Let , then

The integral becomes   

⇒

⇒

Hence option 2 is correct

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Integration by Substitution:

If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.

.

Calculation:

Let I = .

We substitute cos^-1 x = t ⇒ x = cos t and  = dt.

⇒ I = - ∫ t cos t dt

Integrating by parts, we get:

⇒ I = -t ∫ cos t + ∫ (1 × ∫ cos t dt) dt

⇒ I = -t sin t + ∫ sin t dt + C

⇒ I = - t sin t - cos t + C

= I = 

Concept:

• Integration by Parts:

  ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

• ∫ sin x dx = - cos x + C

Calculation:

Let I = ​​∫ ex(sin x - cos x) dx.

⇒ I = ∫ ex sin x dx - ∫ ex cos x dx

⇒ I = ex ∫ sin x dx - ∫ [e^x ∫ sin x dx] dx - ∫ ex cos x dx

⇒ I = - ex cos x dx + ∫ ex cos x dx - ∫ ex cos x dx + C

⇒ I = - ex cos x + C

As we know, ∫ ex [f(x) + f'(x)]dx = ex f(x) + c

Let f(x) = -cos x

So, f'(x) = sin x

Now, I =  ∫ ex(sin x - cos x) dx

=  ​​∫ ex(- cos x + sin x) dx

= ∫ ex [f(x) + f'(x)]dx

=  ex f(x) + c

= - ex cos x + C

Concept:

1. Integration by Substitution:

• If the given integration is of the form  where  and  are both differentiable functions then we substitute  which implies that .
• Therefore, the integral becomes  which can be solved by general formulas.

Solution:

In the given problem substitute  therefore, .

The given integral becomes

Resubstitute .

Concept:

Integration by parts:

The formula for integrating by parts is given by;

Where u is the function u(x) and v is the function v(x)

• ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

In the given function u = x and v = ln xdx.

Integrate by parts as follows:

Concept Used:

1. a logx = log (x^a)

2. e^log(n)  = n

3. ∫xⁿdx = 

Application:

We have,

I = e5 log x

or, I =  = x⁵

Hence, ∫x⁵ dx = x⁶/6 + C

Formula:

Calculation:

Let 

⇒ 

⇒ 

⇒ 

The above integrand is of the form 

Concept:

Integration by parts

Using ILATE rules

I → Inverse function

L → Log function

A → Algebraic function

T → Trigonometry function

E → Exponential function

Calculation:

= 

= 

= 

Put 1 – x² = t²

⇒ -2x dx = 2t dt

⇒ x dx = -t dt

Now, I =  

= 

=  + c

Putting the value of t ⇒ 

I = 

Concept-

Integration by parts formula-: 

Calculation-

Taking log x as first function x as second and using by parts formula

= 

= 

= 

= 

∴ 

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Definite Integral:

If ∫ f(x) dx = g(x) + C, then  = g(b) - g(a).

Calculation:

Let's first integrate the expression under integral by parts.

I = ∫ log x dx =  ∫ (1)(log x) dx

Considering log x as the first function and 1 as the second function, we get:

= (log x) ∫ 1 dx - ∫ [ ∫ 1 dx] dx

= (log x) x - x + C

Putting the limits of the definite integral, we get:

= 

= (2 log 2 - 2) - (0 - 1)

= 2 log 2 - 1.

Concept:

Integration by parts: Integration by parts is a method to find integrals of products.

The formula for integrating by parts is given by:

⇒  + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preferred order for this rule and is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

I = 

I = 

I = 

I = 

I = 

I =