Difficult Integral

Question

I have a problem with solving this integral: $$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$ I tried to use substitution but I got stuck. Can anyone help me?

Answer 1

The easiest way is using undetermined coefficients: $$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$

You must differentiate and you have: $$\frac{2x^2+3x+1}{\sqrt{x^2+1}}=A\sqrt{x^2+1}+(Ax+B)\frac{2x}{2\sqrt{x^2+1}}+\frac{\lambda}{\sqrt{x^2+1}}$$

So you have an equation: $$2x^2+3x+1=A(x^2+1)+(Ax+B)x+\lambda$$ $$2x^2+3x+1=Ax^2+A+Ax^2+Bx+\lambda$$ from which you can find out an simultaneous equation:

\begin{cases} 2=2A\\ 3=B\\ 1=A+\lambda \end{cases}

After solving it you have: \begin{cases} A=1\\B=3\\\lambda=0 \end{cases}

To remind, your integral grom the beginning is:

$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$ So

$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (x+3)\sqrt{x^2+1}$$

Answer 2

Here is another way of doing this integral:

Let $x=\sinh u\Rightarrow dx=\cosh du$. Also $\sqrt{x^2+1}=\cosh u$.

Then $$I=\int (2\sinh^2u+2\sinh u+1 )du$$ $$=\int(\cosh 2u+3\sinh u) du$$ $$=\frac 12 \sinh2u+3\cosh u +c$$ $$=(\sinh u+3)\cosh u+c$$ $$=(x+3)\sqrt{x^2+1}+c$$

Answer 3

Let $y=\arctan x,x=\tan y\implies\sec y=+\sqrt{x^2+1}$ as $-\dfrac\pi2\le y\le\dfrac\pi2$

$$\int\dfrac{2x^2+3x+1}{\sqrt{x^2+1}}dx=\int(2\tan^2y+3\tan y+1)\sec y\ dy$$

$$=2\int\sec^3y\ dy-\int\sec y\ dy+3\int\tan y\ dy$$

Utilize Indefinite integral of secant cubed and replace the values of $\sec y,\tan y$

Answer 4

$$\int\frac{2x^2+3x+1}{\sqrt{x^2+1}}\space\text{d}x=$$

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For the integrand $\frac{2x^2+3x+1}{\sqrt{x^2+1}}$, substitute $x=\tan(u)$ and $\text{d}x=\sec(u)\space\text{d}u$.

Then $\sqrt{x^2+1}=\sqrt{\tan^2(u)+1}=\sec(u)$ and $u=\arctan(u)$:

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$$\int\left(2\tan^2(u)+3\tan(u)+1\right)\sec(u)\space\text{d}u=$$ $$\int\left(\sec(u)+2\tan^2(u)\sec(u)+3\tan(u)\sec(u)\right)\space\text{d}u=$$ $$\int\sec(u)\space\text{d}u+\int 2\tan^2(u)\sec(u)\space\text{d}u+\int3\tan(u)\sec(u)\space\text{d}u=$$ $$\int\sec(u)\space\text{d}u+2\int \tan^2(u)\sec(u)\space\text{d}u+3\int\tan(u)\sec(u)\space\text{d}u=$$ $$2\int\sec^3(u)\space\text{d}u-\int\sec(u)\space\text{d}u+3\int\tan(u)\sec(u)\space\text{d}u=$$

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Use the reduction formula:

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$$\tan(u)\sec(u)+3\int\tan(u)\sec(u)\space\text{d}u=$$

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Substitute $s=\sec(u)$ and $\text{d}s=\tan(u)\sec(u)\space\text{d}u$:

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$$\tan(u)\sec(u)+3\int1\space\text{d}s=$$ $$\tan(u)\sec(u)+3s+\text{C}=$$ $$\tan(u)\sec(u)+3\sec(u)+\text{C}=$$ $$\tan\left(\arctan(x)\right)\sec\left(\arctan(x)\right)+3\sec\left(\arctan(x)\right)+\text{C}=$$ $$(x+3)\sqrt{x^2+1}+\text{C}$$

Answer 5

You can also use one of the Euler's substitutions: $$\begin{align} \sqrt{x^2+1} &= t-x \\ x &= \frac{1}{2}\left(t - \frac{1}{t}\right) \\ dx &= \frac{1}{2}\left(1 + \frac{1}{t^2}\right)dt \\ \sqrt{x^2+1} &= \frac{1}{2}\left(t + \frac{1}{t}\right) \\ \frac{dx}{\sqrt{x^2+1}}&=\frac{dt}{t}\\ 2x^2+3x+1&= \frac{t^4+3t^3-3t+1}{2t^2} \end{align} $$ Integration is now straightforward $$\begin{align}\int\frac{t^4+3t^3-3t+1}{2t^3}dt&=\frac{-t + 6t + 6 t^3 + t^4}{4 t^2}+C\\&=\frac{(t^2+1)(t^2+6 t-1)}{4 t^2}+C \\ &=\frac{1}{2}\left(1+\frac{1}{t}\right)\left[\frac{1}{2}\left(t-\frac{1}{t}\right)+3\right]\\ &=\sqrt{x^2+1}(x+3) \end{align}$$ As it often happens with the Euler's method, the algebra is tedious, but it gets you there.

Answer 6

HINT:

Let $$\dfrac{2x^2+3x+1}{\sqrt{x^2+1}}=\dfrac{2(x^2+1)+3x-1}{\sqrt{x^2+1}}=2\sqrt{x^2+1}+\dfrac32\cdot\dfrac{2x}{\sqrt{x^2+1}}-\dfrac1{\sqrt{x^2+1}}$$

Use this

Answer 7

$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx=\int{\frac{2\sinh^2(t)+3\sinh(t)+1}{\sqrt{\sinh^2(x)+1}}} \cosh(t)dt= \int(\cosh(2t)+3\sinh(t))dt=\frac12\sinh(2t)+3\cosh(t)=x\sqrt{x^2+1}+3\sqrt{x^2+1}$$