## Integration Worksheet5 solution1

In this page integration worksheet5 solution1 we are going to see solution of some practice question from the worksheet of integration.

Question 1

Integrate the following with respect to x,x⁵ (1 + x⁶)⁷

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "1 + x⁶"

t = 1 + x⁶

differentiate with respect to x

dt = 6 x⁵ dx

dt/6 = x⁵ dx

x⁵ dx = dt/6

= ∫ x⁵ (1 + x⁶)⁷ dx

= ∫ t⁷ (dt/6)

= (1/6) t⁷ dt

= (1/6) [t^(7+1)/(7+1)] + C

= (1/6) (t⁸/8) + C

= (1/48) t⁸ + C

= t⁸/48 + C

= (1 + x⁶)⁸/48 + C

Question 2

Integrate the following with respect to x,(2Lx + m)/(Lx² + mx + n)

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "(Lx² + mx + n)"

t = (Lx² + mx + n)

differentiate with respect to x

dt = (2 L x + m (1) + 0) dx

dt = (2 L x + m) dx

now we are going to apply the value of t and dt in the given question

= ∫ (2Lx + m)/(Lx² + mx + n) dx

= ∫ (dt/t)

= log t + C

= log (Lx² + mx + n) + C

Question 3

Integrate the following with respect to x,(4ax + 2b)/(ax² + bx + c)^10

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "(ax² + bx + c)"

t = (ax² + bx + c)

differentiate with respect to x

dt = (2 a x + b (1) + 0) dx

dt = (2 a x + b) dx

now we are going to apply the value of t and dt in the given question

= ∫(4ax + 2b)/(ax² + bx + c)^10 dx

now we are going to take 2 from the numerator

= ∫ 2 (2 ax + 2b)/(ax² + bx + c)^10 dx

= ∫ 2 (dt/t^10)

= ∫ 2 t^-10 dt

= 2 t^(-10 + 1)/(-10 + 1) + C

= 2 t^(-9)/(-9) + C

= (-2/9) (ax² + bx + c)^(-9) + C

= [-2/9(ax² + bx + c)^9] + C