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Evaluate $\int \sqrt{1+e^{2 x}} d x$

let $t=\sqrt{1+e^{2 x}} \longrightarrow t^{2}=1+e^{2 x} \longrightarrow 2 t d t=2 e^{2 x} d x \rightarrow d x=\frac{2td t}{2 e^{2 x}}$

$d x=t \cdot e^{-2 x} d t$

$\int \sqrt{1+e^{2 x}} d x=\int t \cdot t \cdot e^{-2 x} d t=\int \frac{t^{2} d t}{e^{2 x}}$

$t^{2}=1+e^{2 x}, e^{2 x}=t^{2}-1$

$\int \sqrt{1+e^{2 x}} d x=\int \frac{t^{2} d t}{t^{2}-1}$

long division

$=\int\left(1+\frac{1}{t^{2}-1}\right) d t$

$=\int 1 d t+\int \frac{1}{t^{2}-1} d t=t+\int \frac{1}{(t-1)(t+1)}dt$

$\int \frac{1}{(t-1)(t+1)}=\int \frac{A}{t-1} d t+\int \frac{B}{t+1} d t$

$A=(t-1) \cdot\left.\frac{1}{(t-1)(t+1)}\right|_{t=1}=\frac{1}{t+1} |_{t=1}=\frac{1}{2}$

$B=(t+1) \cdot\left.\frac{1}{(t-1)(t+1)}\right|_{t=-1}=\frac{1}{t-1} |_{t=-1}=-\frac{1}{2}$

$\int \frac{1 d t}{(t-1)(t+1)}=\int \frac{\frac{1}{2}}{t-1} d t+\int \frac{-\frac{1}{2}}{t+1} d t$

$=\frac{1}{2} \int \frac{1}{t-1} d t-\frac{1}{2} \int \frac{1}{t+1} d t$

$\int \frac {1}{(t-1)(t+1)}dt=\frac{1}{2} \ln |t-1|-\frac{1}{2}|n| t+1 |+c$

$\int \sqrt{1+e^{2 x}} d x=\int 1 d t+\int \frac{1 d t}{(t-1)(t+1)}$

$\int \sqrt{1+e^{2 x}} d x=t+\frac{1}{2} \ln |t-1|-\frac{1}{2} \ln |t+1|+c$

$\begin{array}{l}{a \ln |{b}|=\ln |b^a|} \\ {\ln |a|-\ln |b|=\ln \left|\frac{a}{b}\right|}\end{array}$

$\int \sqrt{1+e^{2 x}} d x=t+\ln \left|(t-1)^{\frac{1}{2}}\right|-\ln \left|(t+1)^{\frac{1}{2}}\right|+c$

$\int \sqrt{1+e^{2 x}} d x=t+\ln |\sqrt{(t-1)}|-\ln |\sqrt{(t+1)}|+c$

$\int \sqrt{1+e^{2 x}} d x=t+\ln \left|\frac{\sqrt{t-1}}{\sqrt{t+1}}\right|+c$

$t=\sqrt{1+e^{2 x}}$

$\int \sqrt{1+e^{2 x}}=\sqrt{1+e^{2 x}}+\ln \left|\sqrt{\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+ e^{2x}}+1}}\right|+c$

Evaluate $\int \frac{d x}{x^{1 / 2}+x^{1 / 3}}$

$2,3 \rightarrow 6$

let $u=x^{\frac{1}{6}} \Rightarrow x=u^{6} \Rightarrow d x=6 u^{5} d u$

$x^{\frac{1}{2}}=u^{3} \qquad x^{\frac{1}{3}}=u^{2}$

$\int \frac{{6} u^{5} d u}{u^{3}+u^{2}}=6 \int \frac{u^{5}}{u^{3}+u^{2}} d u$

$=6 \int \frac{u^{2}\cdot u^{3}}{u^{2}(u+1)} d u=6 \int \frac{u^{3}}{u+1} d u$

$\int \frac{d x}{x^{1 / 2}+x^{1 / 3}}=6 \int \frac{u^{3}}{u+1} d u=6 \int u^{2}-u+1-\frac{1}{u+1}$

$\int \frac{d x}{x^{1/2}+x^{1/3}}=6\left[\frac{u^{3}}{3}-\frac{u^{2}}{2}+u-\ln |u+1|\right]+c$

$\int \frac{d x}{x^{1 / 2}+x^{{1}/{3}}}=2 u^{3}-3 u^{2}+6 u-\ln |u+1|+c$

$\int \frac{d x}{x^{1/2}+x^{1/3}}=2 x^{\frac{1}{6} \cdot 3}-3 x^{\frac{1}{6}\cdot 2}+6 x^{\frac{1}{6}}-\ln \left|x^\frac{1}{6}+1\right|+c$

$\int \frac{d x}{x^{1/2}+x^{1/3}}=2 \sqrt{x}-3 x^\frac {1}{3}+6 x^\frac {1}{6}-\ln \left|x^\frac {1}{6}+1\right|+C$