Question

$\int g\left(x\right)dx=F\left(x\right)$, then $\int x^{3}g\left(x^2\right)dx=$

• A. $\frac{1}{2}\left[x^2\right(F\left(x\right){)}^2-\int \left(F\right(x{)}^2)dx]$
• B. $\frac{1}{2}\left[x^{2}F\right(x{)}^2-\int F\left(x{)}^{2}d\right(x^2)]$
• C. $\frac{1}{2}\left[x^{2}F\right(x)-\frac{1}{2}\int \left(F\right(x){)}^{2}dx]$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{2}\left[x^{2}F\right(x{)}^2-\int F\left(x{)}^{2}d\right(x^2)]$

$\begin{array}{c}\int g\left(x\right)dx=f\left(x\right),\int x^{3}g\left(x^2\right)dx\\ let{x}^2=t\\ \Rightarrow 2xdx=dt,dx=\frac{dt}{2x}\\ \Rightarrow \frac{1}{2}\int tg\left(t\right)=t\int g\left(t\right)d-\int [\frac{dt}{dt}.\int g\left(t\right)dt]dt\\ \Rightarrow \frac{1}{2}\int g\left(t\right)=tf\left(t\right)-\int F\left(t\right)dt=\frac{1}{2}\left[x^{2}F\left(x^2\right)-\int F\left(x^2\right)-d{x}^2\right].\end{array}$