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प्रश्न
विकल्प
2 loge cos (xex) + C
sec (xex) + C
tan (xex) + C
tan (x + ex) + C
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उत्तर
tan (xex) + C
\[\text{Let }I = \int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)}dx\]
\[\text{Putting }x e^x = t\]
\[ \Rightarrow \left( 1 \cdot e^x + x e^x \right)dx = dt\]
\[ \Rightarrow e^x \left( 1 + x \right)dx = dt\]
\[ \therefore I = \int\frac{dt}{\cos^2 t}\]
\[ = \int \sec^2 t dt\]
\[ = \tan t + C\]
\[ = \tan \left( x e^x \right) + C ............\left( \because t = x e^x \right)\]
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