AP EAMCET · Maths · Indefinite Integration
\(\int(x+1)(x+2)^4(x+3) d x\) is equal to
- A \(\frac{(x+1)^2}{2}+\frac{(x+2)^2}{5}+\frac{(x+3)^2}{2}+C\)
- B \(\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C\)
- C \(\frac{(x+2)^7}{7}+\frac{(x+2)^5}{5}+C\)
- D \(\frac{(x+3)^7}{7}-\frac{(x+3)^5}{5}+C\)
Answer & Solution
Correct Answer
(B) \(\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C\)
Step-by-step Solution
Detailed explanation
Let \(x+2=t\), then \(d x=d t\) \[ \begin{aligned} I & =\int(t-1)(t)^4(t+1) d t=\int\left(t^2-1\right) t^4 d t \\ & =\int\left(t^6-t^4\right) d t \\ & =\frac{t^7}{7}-\frac{t^5}{5}+C \\ & =\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C \end{aligned} \]
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