Evaluate the integral. ∫x2−5x+16(2x+1)(x−2)2dx

Question

Evaluate the integral.  ∫x2−5x+16(2x+1)(x−2)2dx

Toni Scott · Accepted Answer

Step 1 The integral given is, ∫x2−5x+16(2x+1)(x−2)2dx. After partial fraction decomposition, ∫x2−5x+16(2x+1)(x−2)2dx=∫32x+1dx−∫1x−2dx+∫2(x−2)2dx. Solve first part of integral , Substitute u⇒2x+1,du=2dx. ∫32x+1dx=32∫duu =32ln⁡u+c1. Step 2 After replace u with 2x+1, ∫32x+1dx=32ln⁡(2x+1)+c1...(1) Solving the second part of integral, Substitute uRightarrow x-2, du=dx. ∫dxx−2=∫duu =ln⁡u+c2. Replace u with x-2, ∫dxx−2=ln⁡(x−2)+c2...(2) Step 3 Solving the third part of the integral, Substitute u ⇒ x-2, du=dx. ∫dx(x−2)2=∫duu2 =−1u+c3. Replace u with x-2, ∫dx(x−2)2=−1x−2+c3...(3) Step 4 The final solution of the integration becomes, adding equation 1 , 2 and 3, ∫x2−5x+16(2x+1)(x−2)2dx=∫32x+1dx−∫1x−2dx+∫2(x−2)2dx.

Debbie Moore · Accepted Answer

∫x2−5x+16(2x+1)×(x−2)2dx  ∫32x+1−1x−2+2(x−2)2dx  ∫32x+1dx−∫1x−2dx+∫2(x−2)2dx  32×ln⁡(|2x+1|)−ln⁡(|x−2|)−2x−2  Add C  Solution  32×ln⁡(|2x+1|)−ln⁡(|x−2|)−2x−2+C

karton · Accepted Answer

∫x2−5x+16(x−2)2(2x+1)dx=∫(32x+1−1x−2+2(x−2)2)dx=3∫12x+1dx−∫1x−2dx+2∫1(x−2)2dx∫12x+1dx=12∫1udu∫1udu=ln⁡(u)12∫1udu=ln⁡(u)2=ln⁡(2x+1)2∫1x−2dx=∫1udu=ln⁡(u)=ln⁡(x−2)∫1(x−2)2dx=∫1u2du=−1u=−1x−23∫12x+1dx−∫1x−2dx+2∫1(x−2)2dx=3ln⁡(2x+1)2−2x−2−ln⁡(x−2)∫x2−5x+16(x−2)2(2x+1)dx=3ln⁡(|2x+1|)2−2x−2−ln⁡(|x−2|)+C

Evaluate the integral. \int \frac{x^{2}-5x+16}{(2x+1)(x-2)^{2}}dx

Answered question

Answer & Explanation

New Questions in Calculus 2