\(\displaystyle\int x\,(x^2 + 1)^4\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{(x^2 + 1)^5}{10} + C\)
Let \(t = x^2 + 1\) (substitution).
\(dt = 2x\,dx\)
\( \displaystyle x\,dx = \frac{dt}{2} \)
Then
\( \displaystyle \int x\,(x^2 + 1)^4\,dx = \frac{1}{2}\int t^4\,dt \)
\( \displaystyle = \frac{1}{2} \times \frac{t^5}{5} + C \)
\( \displaystyle = \frac{(x^2 + 1)^5}{10} + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{(x^2 + 1)^5}{5} + C\) — Incorrect. The factor \(\frac{1}{2}\) from \(x\,dx = \frac{dt}{2}\) has been left out.
Option (2) \(\displaystyle\frac{(x^2 + 1)^5}{2} + C\) — Incorrect. The division by the new power 5 has been left out.
Option (3) \(\displaystyle\frac{x^2(x^2 + 1)^5}{10} + C\) — Incorrect. The extra factor \(x^2\) should not be there; its derivative brings in further terms that are not in the integrand.