Module 3: Now You Try I

Compute the following integrals.

1. $\int \frac{x^{2}}{\sqrt{1 + 2 x}} d x$ $\int \frac{x^2}{\sqrt{1+2x}} dx$

Solution:

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Let's try

u = 1 + 2 x .

$u=1+2x.$ Then

x = \frac{u - 1}{2}

$x=\frac{u-1}{2}$ and

d u = 2 d x

$du=2dx$ .

\begin{aligned} \int \frac{x^{2}}{\sqrt{1 + 2 x}} d x & = \int \frac{{(\frac{u - 1}{2})}^{2}}{\sqrt{u}} \frac{d u}{2} \\ = \frac{1}{2} \int \frac{u^{2} - 2 u + 1}{4 u^{\frac{1}{2}}} d u \\ = \frac{1}{8} \int (u^{\frac{3}{2}} - 2 u^{\frac{1}{2}} + u^{\frac{- 1}{2}}) d u \\ = \frac{1}{8} (\frac{1}{\frac{3}{2} + 1} u^{\frac{3}{2} + 1} + (- 2) \frac{1}{\frac{1}{2} + 1} u^{\frac{1}{2} + 1} + \frac{1}{\frac{- 1}{2} + 1} u^{\frac{- 1}{2} + 1}) + C \\ = \frac{1}{8} (\frac{1}{\frac{3}{2} + 1} (1 + 2 x)^{\frac{3}{2} + 1} + (- 2) \frac{1}{\frac{1}{2} + 1} (1 + 2 x)^{\frac{1}{2} + 1} + \frac{1}{\frac{- 1}{2} + 1} (1 + 2 x)^{\frac{- 1}{2} + 1}) + C \end{aligned}

$\begin{align*} \int \frac{x^2}{\sqrt{1+2x}} dx&=\int \frac{\left(\frac{u-1}{2}\right)^2}{\sqrt{u}} \frac{du}{2}\\ &=\frac{1}{2}\int \frac{u^2-2u+1}{4u^{\frac{1}{2}}}\; du\\ &=\frac{1}{8}\int \left(u^{\frac{3}{2}}-2u^{\frac{1}{2}}+u^{\frac{-1}{2}}\right)\; du\\ &=\frac{1}{8} \left( \frac{1}{\frac{3}{2}+1}u^{\frac{3}{2}+1}+(-2)\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}+\frac{1}{\frac{-1}{2}+1}u^{\frac{-1}{2}+1}\right) +C\\ &=\frac{1}{8} \left( \frac{1}{\frac{3}{2}+1}(1+2x)^{\frac{3}{2}+1}+(-2)\frac{1}{\frac{1}{2}+1}(1+2x)^{\frac{1}{2}+1}+\frac{1}{\frac{-1}{2}+1}(1+2x)^{\frac{-1}{2}+1}\right) +C\\ \end{align*}$

2. $\int \frac{1}{x \ln x} d x$ $\int \frac{1}{x\ln x} dx$ .

Solution:

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Let

u = \ln x

$u=\ln x$ , then

d u = \frac{1}{x} d x

$du=\frac{1}{x}dx$ .

\begin{aligned} \int \frac{1}{x \ln x} d x & = \int \frac{1}{x u} x d u \\ = \int \frac{1}{u} d u \\ = \ln | u | + C \\ = \ln (| \ln x |) + C . \end{aligned}

$\begin{align*} \int \frac{1}{x\ln x} dx&=\int \frac{1}{xu} xdu\\ &=\int \frac{1}{u} du\\ &=\ln|u|+C\\ &=\ln\left(|\ln x\right|)+C. \end{align*}$