Resolved: "Evaluate these integrals. ∫ex+1dx"

Yasmin

Answered question

2020-11-06

Evaluate these integrals.

\int \sqrt{e^{x} + 1} d x

Answer & Explanation

Delorenzoz

Skilled2020-11-07Added 91 answers

Evaluate:
$\int \sqrt{e^{x} + 1} d x$ (1)
Simplification:
Substitute:
$e^{x} + 1 = t$
$e^{x} d x = d t$
$d x = \frac{d t}{e^{x}}$
$d x = \frac{d t}{t - 1}$
in(1),
$\int \sqrt{e^{x} + 1} d x = \int \frac{\sqrt{t}}{t - 1} d t$ (2)
Now again substitute
$\sqrt{t} = u$
$[\frac{1}{2 \sqrt{t}} d t = d u \Rightarrow d t = 2 \sqrt{t} d u \Rightarrow d t = 2 u^{2} d u]$
in (2),
$\int \sqrt{e^{x} + 1} d x = \int \frac{\sqrt{t}}{t - 1} d t$
$= 2 \int \frac{u}{u^{2} - 1} (2 u) d u$
$= 2 \int \frac{u^{2}}{u^{2} - 1} d u$
$= 2 \int \frac{u^{2} - 1 + 1}{u^{2} - 1} d u$
$= 2 \int (\frac{u^{2} - 1}{u^{2} - 1} + \frac{1}{u^{2} - 1}) d u$
$= 2 \int (1 + \frac{1}{u^{2} - 1}) d u$
$= 2 \int (1 + \frac{1}{(u + 1) (u - 1)}) d u$ (3)
Now evaluate $\frac{1}{(u + 1) (u - 1)}$ using partial fraction method,
$\frac{1}{(u + 1) (u - 1)} = \frac{A}{u + 1} + \frac{B}{u - 1}$
$1 = A (u - 1) + B (u + 1)$
$P u t u = 1, 1 = 2 B \Rightarrow B = \frac{1}{2}$
$P u t u = - 1, 1 = - 2 A \Rightarrow A = - \frac{1}{2}$
$\frac{1}{(u + 1) (u - 1)} = - \frac{1}{2 (u + 1)} + \frac{1}{2 (u - 1)}$ (4)
Now using (4) in (3),
$\int \sqrt{e^{x} + 1} d x = 2 \int (1 + (- \frac{1}{2 (u + 1)} + \frac{1}{2 (u - 1)})) d u$
$= 2 \int (1 - \frac{1}{2 (u + 1)} + \frac{1}{2 (u - 1)}) d u$
$= 2 u - \ln | u + 1 | + \ln | u - 1 | + C$
$= 2 \sqrt{t} - \ln | \sqrt{t} + 1 | + \ln | \sqrt{t} - 1 | + C$
$= 2 \sqrt{e^{x} + 1} - \ln | \sqrt{e^{x} + 1} + 1 | + \ln | \sqrt{e^{x} + 1} - 1 | + C$
Hence
$\int \sqrt{e^{x} + 1} d x = 2 \sqrt{e^{x} + 1} - \ln | \sqrt{e^{x} + 1} + 1 | + \ln | \sqrt{e^{x} + 1} - 1 | + C$

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