If I_1=∫ e^x d x e^x+e^ -x and I_2=∫ d x e^ 2 x +1 , then what is I_1+I_2 equal to?

Correct answer: B. x+c aligned & I_1=∫ e^x e^x+e^ -x d x=∫ (e^x)/(e^x+(1)/(e^x)) d x & =∫ e^ 2 x e^ 2 x+1 d x & I_1+I_2=∫ e^ 2 x +1 e^ 2 x +1 d x=∫ 1 d x & =x+C aligned

NDA 2022

Exam: NDA
Year: 2022
Subject: Mathematics
Chapter: Indefinite Integration

NDA 2022 · Mathematics · Indefinite Integration

If I_1=∫ e^x d x e^x+e^ -x and I_2=∫ d x e^ 2 x +1 , then what is...

I_1=\int \frac{e^x d x}{e^x+e^{-x}}

and

I_2=\int \frac{d x}{e^{2 x}+1}

, then what is

I_1+I_2

equal to?

Options

A $\frac{x}{2}+c$
B $x+c$
C $\ln \left(e^x+e^{-x}\right)+c$
D $\ln \left(e^x-e^{-x}\right)+c$

Correct answer

B. $x+c$

Step-by-step solution

\begin{aligned} & I_1=\int \frac{e^x}{e^x+e^{-x}} d x=\int \frac{e^x}{e^x+\frac{1}{e^x}} d x \\ & =\int \frac{e^{2 x}}{e^{2 x+1}} d x \\ & \therefore I_1+I_2=\int \frac{e^{2 x}+1}{e^{2 x}+1} d x=\int 1 d x \\ & =x+C\end{aligned}

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