# Using substitution to evaluate indefinite integral $\int{x\sqrt{4x+1}}dx$

Evaluate this indefinite integral. $$I= \int{x\sqrt{4x+1}}dx$$

Let $u=4x+1$
$$\frac{du}{dx}=4\rightarrow{dx=\frac{du}{4}}$$ $$I=\int{x}\sqrt{u}\frac{1}{4}du=\frac{1}{4}\int{x}\sqrt{u}du$$

Then I got stuck at this point.

• As soon as you made your change of variable u = 4 x + 1, x must not appear anymore anywhere. I think that this is why you got stuck. – Claude Leibovici Jan 12 '14 at 6:59
• Can you look at the answer down there please and tell me how did he get $u^{3/2}$ in the third step ? – Out Of Bounds Jan 12 '14 at 20:57

Because we have,$$u = 4x +1 \implies u - 1 = 4x \implies \frac{u - 1}{4} = x$$

Can you do the rest?

• $\frac { 1 }{ 4 } \int { \frac { u-1 }{ 4 } } { u }^{ 1/2 }\quad du\\ \\ \frac { 1 }{ 4 } \int { \frac { 1 }{ 4 } } u-\frac { 1 }{ 4 } { u }^{ 1/2 }\quad du\\ \\ \frac { 1 }{ 16 } \int { u } -\frac { 1 }{ 4 } \int { { u }^{ 1/2 } } \quad du\\ \\ \frac { 1 }{ 16 } \times \frac { 1 }{ 2 } { u }^{ 2 }\quad -\quad \frac { 1 }{ 4 } \times \frac { 2 }{ 3 } { u }^{ 3/2 }\quad +\quad C\\ \\ \frac { 1 }{ 32 } { (4x+1) }^{ 2 }\quad -\quad \frac { 1 }{ 6 } { (4x+1 })^{ 3/2 }\quad +\quad C$ – Out Of Bounds Jan 12 '14 at 1:51
• But the answer in the book is $\frac { 1 }{ 40 } { (4x+1) }^{ 5/2 }\quad -\quad \frac { 1 }{ 24 } { (4x+1 })^{ 3/2 }\quad +\quad C$ – Out Of Bounds Jan 12 '14 at 1:52
• @Tennisman: Multiply both terms by $\frac{1}{4}$ in the seind line... – DJohnM Jan 12 '14 at 3:26

$$I= \int{x\sqrt{4x+1}}dx$$

Using Euler Substitution

$$\sqrt{4x+1}=t\iff x=\frac{t^2-1}{4}\iff dx=\frac{t}{2}dt$$

\begin{align} I&= \int{x\sqrt{4x+1}}dx\\ &=\int \left(\frac{t^2-1}{4}\right)\frac{t^2}{2}dt\\ &=\frac{t^5}{40}-\frac{t^3}{24}+C\\ &I=\frac{(4x+1)^{5/2}}{40}-\frac{(4x+1)^{3/2}}{24}+C\\ \end{align}

$$\int{x\sqrt{4x+1}}dx=\frac{3(4x+1)^{5/2}-5(4x+1)^{3/2}}{120}+C$$

$$I = \int x \sqrt{4x+1} \ dx$$

let $u$ = $4x+1$

$x = \frac{u-1}{4}$

$\text{d}x = \frac{\text{d}u}{4}$

\begin{align} I &= \int x \sqrt{4x+1} \ \text{d}x \\ &= \int \left(\frac{u-1}{4} \right) u^{1/2} \frac{\text{d}u}{4} \\ &= \frac{1}{16}\int u^{3/2}\text{d}u - \frac{1}{16}\int u^{1/2}\text{d}u \\ &= \frac{1}{16}\frac{u^{5/2}}{5/2} - \frac{1}{16}\frac{u^{3/2}}{3/2} + \text{C} \\ &= \frac{1}{40}u^{5/2} - \frac{1}{24}u^{3/2} + \text{C} \\ &= \frac{1}{40}(4x+1)^{5/2} - \frac{1}{24}(4x+1)^{3/2} + \text{C} \end{align}

• How did you get $u^{3/2}$ in the third step ? – Out Of Bounds Jan 12 '14 at 2:55
• @Tennisman.Develop what is inside parentheses : u times u^(1/2) = u^(3/2) – Claude Leibovici Jan 13 '14 at 7:22

\begin{align} I &= \int \left(x*\sqrt{4x + 1}\right)dx \\ u &= 4x + 1 \\ (4x + 1)dx &= du \\ \left(\frac{d}{dx}(4x + 1)\right)dx &= du \\ \left(\frac{d}{dx}(4x) + \frac{d}{dx}(1) \right)dx &= du \\ \left(\frac{d}{dx}(4x) + 0 \right)dx &= du \\ \left(4*\frac{d}{dx}(x) + 0\right)dx &= du \\ \left(4*\frac{dx}{dx} + 0\right)dx &= du \\ (4 + 0)dx &= du \\ (4)dx &= du \\ dx &= \left(\frac{1}{4}\right)du \\ I &= \int\left(x*\sqrt{u}*\frac{1}{4}\right)du \\ &= \frac{1}{4}*\int\left(x*\sqrt{u}\right)du \\ u &= 4x + 1 \\ 4x + 1 &= u \\ 4x &= u - 1 \\ x &= \left(\frac{u - 1}{4}\right) \\ I &= \frac{1}{4}*\int\left(\frac{(u - 1)}{4}*\sqrt{u}\right)du \\ &= \frac{1}{4}*\frac{1}{4}*\int\left((u - 1)*\sqrt{u}\right)du \\ &= \frac{1}{16}*\int\left((u - 1)*\sqrt{u}\right)du \\ &= \frac{1}{16}*\int\left(\sqrt{u}*u - \sqrt{u}\right)du \\ &= \frac{1}{16}*\int\left(u^{\frac{1}{2}}*u^1 - u^{\frac{1}{2}}\right)du \\ &= \frac{1}{16}*\int\left(u^{(\frac{1}{2} + 1)} - u^{\frac{1}{2}}\right)du \\ &= \frac{1}{16}*\int\left(u^{(\frac{1}{2} + \frac{2}{2})} - u^{\frac{1}{2}}\right)du \\ &= \frac{1}{16}*\int\left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right)du \\ &= \frac{1}{16}*\left(\int\left(u^{\frac{3}{2}}\right)du - \int\left(u^{\frac{1}{2}}\right)du\right) \\ &= \frac{1}{16}*\left(\int\left(u^{\frac{3}{2}}\right)du - \frac{1}{\left(\frac{1}{2} + 1 \right)}*u^{\left(\frac{1}{2} + 1\right)}\right)\\ &= \frac{1}{16}*\left(\int\left(u^{\frac{3}{2}}\right)du - \frac{1}{\left(\frac{1}{2} + \frac{2}{2}\right)}*u^{\left(\frac{1}{2} + \frac{2}{2}\right)}\right) \\ &= \frac{1}{16}*\left(\int\left(u^{\frac{3}{2}}\right)du - \frac{1}{\left(\frac{3}{2} \right)}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\left(\int\left(u^{\frac{3}{2}}\right)du - \frac{2}{3}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\left(\frac{1}{\left(\frac{3}{2} + 1\right)}*u^{\left(\frac{3}{2} + 1\right)} - \frac{2}{3}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\left(\frac{1}{\left(\frac{3}{2} + \frac{2}{2}\right)}*u^{\left(\frac{3}{2} + \frac{2}{2}\right)} - \frac{2}{3}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\left(\frac{1}{\left(\frac{5}{2}\right)}*u^{\frac{5}{2}} - \frac{2}{3}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\left(\frac{2}{5}*u^{\frac{5}{2}} - \frac{2}{3}*u^{\frac{3}{2}}\right) \\ &= \frac{1}{16}*\frac{2}{5}*u^{\frac{5}{2}} - \frac{1}{16}*\frac{2}{3}*u^{\frac{3}{2}} \\ &= \frac{2}{80}*u^{\frac{5}{2}} - \frac{2}{48}*u^{\frac{3}{2}} \\ &= \frac{1}{40}*u^{\frac{5}{2}} - \frac{1}{24}*u^{\frac{3}{2}} \\ u &= 4x + 1 \\ I &= \frac{1}{40}*\left(4x + 1\right)^{\frac{5}{2}} - \frac{1}{24}*\left(4x + 1\right)^{\frac{3}{2}} \\ I &= \bf\left[\frac{1}{40}*\left(4x + 1\right)^{\frac{5}{2}} - \frac{1}{24}*\left(4x + 1\right)^{\frac{3}{2}} + C \right] \\ \end{align}

## #

Can anyone explain how wolfram gets this result?:

$$\frac{1}{60}*\left(4x+1\right)^{\frac{3}{2}}*\left(6x-1\right) + C$$

• $\displaystyle \frac{1}{40}(4x+1)^{2.5}-\frac{1}{24}(4x+1)^{1.5}=\frac{1}{120}(4x+1)^{1.5} \left[3(4x+1)-5 \right]=\frac{1}{60}(4x+1)^{1.5}(6x-1)$ – Galc127 Nov 23 '14 at 9:37

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int x\root{4x + 1}\,\dd x} ={1 \over 4}\bracks{% \int\pars{4x + 1}^{3/2}\,\dd x - \int\pars{4x + 1}^{1/2}\,\dd x} \\[3mm]&={1 \over 4}\bracks{{\pars{4x + 1}^{5/2} \over 4\times 5/2} -{\pars{4x + 1}^{3/2} \over 4\times 3/2}} =\color{#66f}{\large{3\pars{4x + 1}^{5/2} - 5\pars{4x + 1}^{3/2} \over 120}} + \pars{~\mbox{a constant}~} \end{align}