The value of the integral _-1^1(x^3+|x|+1x^2+2|x|+1)dx is equal to:

Question

The value of the integral $\int _{-1}^{1}(\frac{{x}^{3}+|x|+1}{{x}^{2}+2|x|+1})dx$ is equal to:

Accepted Answer

$I={\int }_{1}^{-1}(\frac{{x}^{3}+|x|+1}{{x}^{2}+2|x|+1})\, dx$

We will use the property of modulus carefully.

Since $|x|$ changes at $x=0$, split the integral into two parts:

$I={\int }_{0}^{-1}(\frac{{x}^{3}+|x|+1}{{x}^{2}+2|x|+1})\, dx+{\int }_{1}^{0}(\frac{{x}^{3}+|x|+1}{{x}^{2}+2|x|+1})\, dx$

For $-1\le x
Here, $|x|=-x$.

So,

${x}\^{3}+|x|+1={x}\^{3}-x+1$

and

${x}\^{2}+2|x|+1={x}\^{2}-2x+1=(x-1{)}\^{2}$

Thus,

${I}\_{1}={\int }\_{0}\^{-1}\frac{{x}\^{3}-x+1}{(x-1{)}\^{2}}\, dx$

For $0\le x\le 1$
Here, $|x|=x$.

So,

${x}\^{3}+|x|+1={x}\^{3}+x+1$

and

${x}\^{2}+2|x|+1={x}\^{2}+2x+1=(x+1{)}\^{2}$

Thus,

${I}\_{2}={\int }\_{1}\^{0}\frac{{x}\^{3}+x+1}{(x+1{)}\^{2}}\, dx$

Hence,

$I={I}\_{1}+{I}\_{2}$

Now simplify both integrands.

Simplifying ${I}\_{1}$
We divide:

$\frac{{x}\^{3}-x+1}{(x-1{)}\^{2}}=\frac{{x}\^{3}-x+1}{{x}\^{2}-2x+1}$

On division,

$\frac{{x}\^{3}-x+1}{{x}\^{2}-2x+1}=x+2+\frac{2x-1}{(x-1{)}\^{2}}$

Also,

$\frac{2x-1}{(x-1{)}\^{2}}=\frac{2(x-1)+1}{(x-1{)}\^{2}}=\frac{2}{x-1}+\frac{1}{(x-1{)}\^{2}}$

So,

${I}\_{1}={\int }\_{0}\^{-1}(x+2+\frac{2}{x-1}+\frac{1}{(x-1{)}\^{2}})\, dx$

Integrating term by term,

$\int x\, dx=\frac{{x}\^{2}}{2},\, \int 2\, dx=2x$

$\int \frac{2}{x-1}\, dx=2\ln |x-1|$

$\int \frac{1}{(x-1{)}\^{2}}\, dx=\int (x-1{)}\^{-2}\, dx=-\frac{1}{x-1}$

Therefore,

${I}\_{1}={[\frac{{x}\^{2}}{2}+2x+2\ln |x-1|-\frac{1}{x-1}]}\_{0}\^{-1}$

At $x=0$:

$\frac{0}{2}+0+2\ln 1-\frac{1}{-1}=1$

At $x=-1$:

$\frac{1}{2}-2+2\ln 2-\frac{1}{-2}=\frac{1}{2}-2+2\ln 2+\frac{1}{2}=-1+2\ln 2$

So,

${I}\_{1}=1-(-1+2\ln 2)=2-2\ln 2$

Simplifying ${I}\_{2}$
Now,

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

On division,

$\frac{{x}\^{3}+x+1}{(x+1{)}\^{2}}=x-2+\frac{4x+3}{(x+1{)}\^{2}}$

Also,

$\frac{4x+3}{(x+1{)}\^{2}}=\frac{4(x+1)-1}{(x+1{)}\^{2}}=\frac{4}{x+1}-\frac{1}{(x+1{)}\^{2}}$

So,

${I}\_{2}={\int }\_{1}\^{0}(x-2+\frac{4}{x+1}-\frac{1}{(x+1{)}\^{2}})\, dx$

Integrating term by term,

${I}\_{2}={[\frac{{x}\^{2}}{2}-2x+4\ln |x+1|+\frac{1}{x+1}]}\_{1}\^{0}$

At $x=1$:

$\frac{1}{2}-2+4\ln 2+\frac{1}{2}=-1+4\ln 2$

At $x=0$:

$0-0+4\ln 1+1=1$

So,

${I}\_{2}=(-1+4\ln 2)-1=-2+4\ln 2$

Adding ${I}\_{1}$ and ${I}\_{2}$
$I=(2-2\ln 2)+(-2+4\ln 2)=2\ln 2$

Therefore,

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Solutions · 1