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