(B) : \(I=\int x^{3} e^{x^{2}} d x\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 x d x=d t \Rightarrow x \cdot d x=\frac{d t}{2}\) \(\therefore \quad I=\int t \cdot e^{t} \cdot \frac{d t}{2}\) \(=\frac{1}{2}\left[e^{t} t-e^{t}+c\right]=\frac{1}{2}\left[x^{2} e^{x^{2}}-e^{x^{2}}+c\right]\) \(=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
(B) : \(I=\int x^{3} e^{x^{2}} d x\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 x d x=d t \Rightarrow x \cdot d x=\frac{d t}{2}\) \(\therefore \quad I=\int t \cdot e^{t} \cdot \frac{d t}{2}\) \(=\frac{1}{2}\left[e^{t} t-e^{t}+c\right]=\frac{1}{2}\left[x^{2} e^{x^{2}}-e^{x^{2}}+c\right]\) \(=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
(B) : \(I=\int x^{3} e^{x^{2}} d x\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 x d x=d t \Rightarrow x \cdot d x=\frac{d t}{2}\) \(\therefore \quad I=\int t \cdot e^{t} \cdot \frac{d t}{2}\) \(=\frac{1}{2}\left[e^{t} t-e^{t}+c\right]=\frac{1}{2}\left[x^{2} e^{x^{2}}-e^{x^{2}}+c\right]\) \(=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
(B) : \(I=\int x^{3} e^{x^{2}} d x\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 x d x=d t \Rightarrow x \cdot d x=\frac{d t}{2}\) \(\therefore \quad I=\int t \cdot e^{t} \cdot \frac{d t}{2}\) \(=\frac{1}{2}\left[e^{t} t-e^{t}+c\right]=\frac{1}{2}\left[x^{2} e^{x^{2}}-e^{x^{2}}+c\right]\) \(=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
(B) : \(I=\int x^{3} e^{x^{2}} d x\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 x d x=d t \Rightarrow x \cdot d x=\frac{d t}{2}\) \(\therefore \quad I=\int t \cdot e^{t} \cdot \frac{d t}{2}\) \(=\frac{1}{2}\left[e^{t} t-e^{t}+c\right]=\frac{1}{2}\left[x^{2} e^{x^{2}}-e^{x^{2}}+c\right]\) \(=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)