PRACTICE QUESTIONS ON INTEGRATION

Integrate the following with respect to x

(1)  3x³+7x²-2x+1

(2)  (7x²+2x+3)/x⁵

(3)  (x-1) (x+2)

(4)  (x + 1)/x²

(5)  (x + (1/x))² 

(6)  ∫(1 - x)³ 

(7)  (1/x⁵)

(8)  1/x^5/2

(9)  1/∛x⁵ 

(10)  (1/x³)^(1/4)

(11)  

12)

13)  

14)  ∫ [1/x (1 + log x)²] dx

Question 1 :

∫(3x³ + 7x² - 2x + 1) dx

Solution :

=  ∫(3x³ + 7x² - 2x + 1) dx

Now we are going to write separate integration.

=  ∫3x³ dx + ∫7x²dx -∫2x dx + ∫1 dx

=  3∫x³ dx + 7∫x²dx - 2∫xdx + ∫1dx

=  3(x⁴/4) + 7 (x³/3) - 2(x²/2) + x + C

=  3x⁴/4 + 7x³/3 - x² + x + C

So, the answer is 3x⁴/4 + 7x³/3 - x² + x + C.

Question 2 :

∫(7x²+2x+3)/x⁵ dx

Solution :

=  ∫[(7x²+2x+3)/x⁵] dx

Decomposing it, we get

=  ∫(7x²/x⁵) dx + ∫(2x/x⁵) dx + ∫(3/x⁵) dx

=  7∫x^-3 dx + 2∫x^-4 dx + 3∫x^-5 dx

=  (7x^-2)/(-2) + (2x^-3)/(-3) + 3x^-4/(-4) + C

=  -7/2 x² - 2/3x³ - 3/4x⁴ + C

So, the answer is -7/2 x² - 2/3x³ - 3/4x⁴ + C.

Question 3 :

∫(x-1) (x+2) dx

Solution :

=  ∫ (x-1) (x+2) dx

=  ∫(x²+x-2) dx

=  ∫ x² dx + ∫x dx - ∫2 dx

=  (x³/3) + (x²/2) - 2x + C

So, the answer is (x³/3) + (x²/2) - 2x + C.

Question 4 :

∫(x + 1)/x² dx

Solution :

=  ∫(x+1)/x² dx

=  ∫(x/x²) dx+∫(1/x²) dx

=  ∫(1/x) dx + ∫(x⁻²) dx

=  log x - x^-1

=  log x - 1/x + C

So, the answer is log x - 1/x + C.

Question 5 :

∫(x + (1/x))² dx

Solution :

(x + (1/x))²  =  x² + (1/x²) + (2 ⋅ x ⋅ 1/x )

=  ∫x² dx + ∫(1/x²) dx + 2∫dx

=  ∫x² dx + ∫x^-2 dx + 2∫dx

=  (x³/3) - 1/x + 2x + C

So, the answer is (x³/3) - 1/x + 2x + C.

Question 6 :

∫(1 - x)³ dx

Solution :

(1 - x)³  =  1³ - 3(1)²x + 3x² - x³

=  1 - 3x + 3 x² - x³

∫(1 - x)³ dx  =  ∫(1 - 3x + 3x² - x³) dx

=  ∫1 dx - ∫3x dx + ∫3x² dx - ∫x³ dx

=  x-3x²/2 + 3x³/3-x⁴/4 + C

=  x - 3x²/2 + x ³ - x⁴/4 + C

So, the answer is x - 3x²/2 + x³ -  x⁴/4 + C.

Question 7 :

∫ (1/x⁵) dx

Solution :

1/x⁵  =  x^-5

=  x^-4/(-4) + C

=  -1/4x⁴ + C

So, the answer is -1/4x⁴ + C.

Question 8 :

1/x^5/2

Solution :

∫ x^(-5/2) dx  =  x^(-5/2)+1]/(-5/2)+1)

=  x^(-3/2)/(-3/2)+C

=  (-2/3)x^(-3/2)+C

= -2/3x^3/2+C

=  -2/3x√x + C

So, the answer is -2/3x√x + C.

Question 9 :

∫ 1/∛x⁵ dx

Solution :

1/∛x⁵  =  x^(-5/3)

∫ x^(-5/3) dx  = x^(-5/3)+1)/(-5/3)+1

=  x^(-2/3)/(-2/3) + C

=  -3/2x^(2/3) + C

So, the answer is -3/2x^(2/3) + C.

Question 10 :

(1/x³)^(1/4)

Solution :

(1/x³)^(1/4)  =  x^(-3/4)

∫ x^(-3/4) dx  =  x^(-3/4)+1)/(-3/4)+1)

=  x^(1/4)/(1/4) + C

=  4x^(1/4) + C

So, the answer is 4x^(1/4) + C.

Question 11 :

Solution :

Since the upper and lower limit both are same and we have negative sign for lower limit and positive sign for upper limit, we may use the property follows.

Check if the function f(x) is odd or even.

• If it is odd, the answer will become 0.
• If it is even, then 

Let f(x) = x √8 -x²

f(-x) = -x√8 -(-x)²

= -x√8 -x²

Since f(-x) = -f(x), it is even function, then the answer is 0

Question 12 :

Solution :

Using substitution method, let us solve this problem

Let t = x² + 1

x² = t - 1

Find the derivative on both sides,

2x dx = dt - 0

x dx = dt/2

Question 13 :

Solution :

Using substitution method, we can try to do this problem

Let t = x² + 2

Differentiating with respect to x on both sides.

dt = 2x dx

When x = 1, t = 1² + 2 ==> 3

When x = 2, t = 2² + 2 ==> 6

Question 14 :

∫ [1/x (1 + log x)²] dx

Solution :

Let us use substitution method to solve the problem.

Let 1 + log x = t

0 + (/x) dx = dt

∫ [1/x (1 + log x)²] dx = ∫ (1/t²) dt

=  ∫t^-2 dt

= t^-2+1 / (-2 + 1) + C

= t^-1 / (-1) + C

= -1/t + C

Applying the value of t, we get

= -1/(1 + log x) + C

Question 15 :

∫ (6x + 4) / (x - 2)(x - 3) dx

Solution :

∫ (6x + 4) / (x - 2)(x - 3) dx

Since the linear factors are multiplying at the denominator, using the concept of partial fraction we have to decompose into two fractions.

(6x + 4) / (x - 2)(x - 3) = A/(x - 2) + B/(x - 3)  ------(1)

6x + 4 = A(x - 3) + B(x - 2)

Put x = 3

6(3) + 4 = A(3 - 3) + B(3 - 2)

18 + 4 = B

B = 22

Put x = 2

6(2) + 4 = A(2 - 3) + B(2 - 2)

12 + 4 = A(-1)

A = -16

Applying the values of A and B in (1), we get

(6x + 4) / (x - 2)(x - 3) = -16/(x - 2) + 22/(x - 3)

Integrating on both sides,

∫ (6x + 4) / (x - 2)(x - 3) dx = -∫16/(x - 2) dx + ∫22/(x - 3) dx

= -16log (x - 2) + 22 log (x - 3) + C

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