1 ∫ 0 E X 1 + E 2 X D X

प्रश्न

\[\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx\]

उत्तर

\[Let\ e^x = t . Then, e^x dx = dt\]
\[When\ x = 0, t = 1\ and\ x = 1, t = e\]
\[ \therefore I = \int_0^1 \frac{e^x}{1 + e^{2x}} d x\]
\[ \Rightarrow I = \int_1^e \frac{dt}{1 + t^2}\]
\[ \Rightarrow I = \left[ \tan^{- 1} x \right]_1^e \]
\[ \Rightarrow I = \tan^{- 1} e - \tan^{- 1} 1\]
\[ \Rightarrow I = \tan^{- 1} e - \frac{\pi}{4}\]

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Definite Integrals

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 6 Q 5 Q 7

अध्याय 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12

अध्याय 19 Definite Integrals
Exercise 20.2 | Q 6 | पृष्ठ ३८

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