Question

$\int x^{27}\left(6 x^2+5 x+4\right)\left(1+x+x^2\right)^6 d x$ is equal to

• A. $\frac{x^4}{7} \cdot\left(1+x+x^2\right)^7$
• B. $\frac{ x ^{28}\left(1+ x + x ^2\right)^7}{7}+C$
• C. $\frac{x^{28}\left(1+x+x^2\right)^7}{28}+C$
• D. $\frac{x^{28}\left(1+x+x^2\right)^6}{6}+C$

Answer 1

Answer: (B)

$I=\int\left(6 x^5+5 x^4+4 x^3\right)\left(x^4+x^5+x^6\right)^6 d x$
Let $x ^6+ x ^5+ x ^4= t$
$\left(6 x^5+5 x^4+4 x^2\right) d x=d t$
$I=\int t^6 d t=\frac{\left(x^4+x^5+x^6\right)^7}{7}+C=\frac{x^{28}\left(1+x+x^2\right)^7}{7}+C$