Question

$\int x^3\left(x^2-1\right)^{1 / 4} d x$

• A. $ \frac{2\left( x ^2-1\right)^{5 / 4}}{45}\left(4+5 x ^2\right)+ C$
• B. $\frac{2\left( x ^2-1\right)^{5 / 4}}{45}\left(5+4 x ^2\right)+ C$
• C. $\frac{\left(x^2-1\right)^{5 / 4}}{45}\left(4+x^2\right)+C$
• D. $\frac{\left(x^2-1\right)^{4 / 5}}{45}\left(4 x^2+1\right)+C$

Answer 1

Answer: (A)

$I=\int x^3\left(x^2-1\right)^{1 / 4} d x$
Hint: $x ^2-1= t ^4 ; xdx =2 t ^3 dt$
$I =\int 2\left(t^4+1\right) t \cdot t^3 d t $
$=2 \cdot \int\left(t^8+t^4\right) d t=\frac{2}{9} t^9+\frac{2 t^5}{5}=\frac{2}{9}\left(x^2-1\right)^{9 / 4}+\frac{2}{5}\left(x^2-1\right)^{5 / 4} $
$ =\frac{2}{45}\left(x^2-1\right)^{5 / 4}\left(5 x^2+4\right) \Rightarrow (A)$