How do you evaluate the integral $\int \sqrt{e^{x} + 1}$ $int sqrt(e^x+1)$ ?

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How do you evaluate the integral $\int \sqrt{e^{x} + 1}$ $int sqrt(e^x+1)$ ?

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Answer 1 · 2018-05-23T06:28:24

Use the substitution $\sqrt{e^{x} + 1} = u$ $sqrt(e^x+1)=u$ .

Explanation:

Let

$I = \int \sqrt{e^{x} + 1} d x$ $I=intsqrt(e^x+1)dx$

Apply the substitution $\sqrt{e^{x} + 1} = u$ $sqrt(e^x+1)=u$ :

$I = 2 \int \frac{u^{2}}{u^{2} - 1} d u$ $I=2int(u^2)/(u^2-1)du$

Rearrange:

$I = 2 \int (1 + \frac{1}{u^{2} - 1}) d u$ $I=2int(1+1/(u^2-1))du$

Factorize the denominator:

$I = 2 \int (1 + \frac{1}{(u - 1) (u + 1)}) d u$ $I=2int(1+1/((u-1)(u+1)))du$

Apply partial fraction decomposition:

$I = \int (2 + \frac{1}{u - 1} - \frac{1}{u + 1}) d u$ $I=int(2+1/(u-1)-1/(u+1))du$

Integrate term by term:

$I = 2 u + ln | u - 1 | - ln | u + 1 | + C$ $I=2u+ln|u-1|-ln|u+1|+C$

Reverse the substitution:

$I = 2 \sqrt{e^{x} + 1} + ln ∣ ∣ ∣ \frac{\sqrt{e^{x} + 1} - 1}{\sqrt{e^{x} + 1} + 1} ∣ ∣ ∣ + C$ $I=2sqrt(e^x+1)+ln|(sqrt(e^x+1)-1)/(sqrt(e^x+1)+1)|+C$

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How do you evaluate the integral $\int \sqrt{e^{x} + 1}$ $int sqrt(e^x+1)$ ?

1 Answer

Explanation:

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