MHT CET · Maths · Indefinite Integration
\(\int x^{3} \cdot e^{x^{2}} d x=\)
- A \(\frac{1}{2} e^{x^{2}}\left(x^{2}+1\right)+c\)
- B \(\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
- C \(\frac{1}{2} e^{x}\left(x^{2}-1\right)+c\)
- D \(\frac{1}{2} e^{x}\left(x^{2}+1\right)+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int x^{3} e^{x^{2}} d x\)
Put \(x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t\)
\(\therefore \mathrm{I}=\frac{1}{2} \int \mathrm{te}^{\mathrm{t}} \mathrm{dt}\)
\(=\frac{1}{2}\left[t e^{t}-e^{t}\right]+c=\frac{1}{2} e^{t}(t-1)+c=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
Put \(x^{2}=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{1}{2} d t\)
\(\therefore \mathrm{I}=\frac{1}{2} \int \mathrm{te}^{\mathrm{t}} \mathrm{dt}\)
\(=\frac{1}{2}\left[t e^{t}-e^{t}\right]+c=\frac{1}{2} e^{t}(t-1)+c=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c\)
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