Evaluating integral $\int\frac{x+1}{\sqrt{x^2+2x+3}}dx$

Question

I have a reoccurring problem when doing the following type of problem.

$$\frac{x+1}{\sqrt{x^2+2x+3}}$$ and: $$\frac{4x-2}{\sqrt{x^2+10x+16}}$$

For some reason, I always end up by dividing in half.

For example, the answer to the first one is: $\sqrt{x^2+2x+3}$ and I calculate $\frac{\sqrt{x^2+2x+3}}{2}$.

Here is how I do it: $$\int\frac{x+1}{\sqrt{x^2+2x+3}}dx$$ Square of polynome $(x+1)^2+2$, then $u=x+1 \to x= u-1$ and $a=\sqrt{2}$ $$\int\frac{u-1+1}{\sqrt{u^2+a^2}}du\to \int \dfrac{u}{\sqrt{u^2+a^2}}du$$ Substitution $w=u^2+a^2$, $\frac{dw}{du}=2u \to du = \frac{1}{2u}dw$ $$\int \dfrac{u}{\sqrt{w}}\dfrac{1}{2u}dw \to \dfrac{1}{2}\int\dfrac{1}{\sqrt{w}}dw \to \dfrac12\int w^{-\frac12} dw = \dfrac12 w^{\frac12}$$ $$\text{Final result} = \dfrac{\sqrt{x^2+2x+2}}{2} \neq \sqrt{x^2+2x+2}$$

I feel like I am missing a point or something. Could someone point out where I keep missing it? Thank you.

Answer 1

Hint: since $$(x^2+2x+3)'=2x+2$$ you can substitute $$t=x^2+2x+3$$ then we get $$dt=2(x+2)dx$$ and $$dx=\frac{1}{2}\frac{1}{x+2}dt$$ And write your second integral in the form $$2\int\frac{2x+10}{\sqrt{x^2+10x+16}}dx-22\int\frac{dx}{\sqrt{x^2+10x+16}}$$

Answer 2

Hint:

For the second integral,, you have first to decompose it into a sum \begin{align} \int\frac{4x-2}{\sqrt{x^2+10x+16}}\,\mathrm d x&= \int\frac{4x+20}{\sqrt{x^2+10x+16}}\,\mathrm d x-\int\frac{22}{\sqrt{x^2+10x+16}}\,\mathrm d x\\&= 2\int\frac{2x+10}{\sqrt{x^2+10x+16}}\,\mathrm d x-22\int\frac{1}{\sqrt{x^2+10x+16}}\,\mathrm d x, \end{align} where the numerator of the first integral is a multiple of the derivative of the radicand and the numerator of the second integral is a constant.

The substitution $u=x^2+10x+16$ yields the first integral. As to the second integral, complete the square in the radicand: $x^2+10x+16=(x+5)^2-9$, and setting $u=x+5$u^, you obtain the integral $$\int\frac{\mathrm du}{u^2-9}$$ which you can calculate decomposing the fraction into simple fractions.

Answer 3

Note that:

$$\int [f(x)]^nf'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$$ whenever $n \ne -1$.

The given integral is

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

This can be rewritten as

$$\int (x+1)(x^2+2x+3)^{-1/2}dx=\frac{1}{2}\int (2x+2)(x^2+2x+3)^{-1/2}dx$$ (RHS is obtained by multiplying and dividing by $2$ in order to use the result mentioned above).

Using the first result mentioned above, we get:

$$\int \frac{x+1}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}\cdot\frac{\left(x^2+2x+3\right)^{1/2}}{\frac{1}{2}}+c=\sqrt{x^2+2x+3}+c$$

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The second problem:

$x^2+10x+16$ can be rewritten as $x^2+10x+25-9=(x+5)^2-3^2$.

Can you take it from here?