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Chapter 1: Problem 21

Evaluate the following integrals: $$ \int \frac{x^{3}+1}{\sqrt{x^{2}+x}} d x $$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{x^{3}+1}{\sqrt{x^{2}+x}} d x$$ Answer: The result of the integral is $$\frac{1}{5}(x^2+x)^\frac{5}{2} - (x^2+x)^\frac{3}{2} + 3\sqrt{x^2+x} + C$$

Step by step solution

01

Simplify the integrand

First, it is helpful to factor out the highest power of \(x\) from the square root, making it easier to work with. In our case, the highest power in the square root is x: $$ \int \frac{x^{3}+1}{\sqrt{x^{2}+x}} d x = \int \frac{x^{3}+1}{\sqrt{x(x+1)}} d x $$
02

Find a suitable substitution

Now, let us perform a substitution, setting \(u = x^2 + x\). Thus, the derivative is \(du = (2x + 1) dx\). Since we have \(x\) terms in the numerator, we need to solve for \(x\) in terms of \(u\). So, let us find \(x\) in terms of \(u\). From the initial substitution, we have \(u - x^2 = x\), which can be rearranged to the quadratic equation \(x^2 + x - u = 0\). Use the quadratic formula to solve for x: $$ x = \frac{-1 \pm \sqrt{1^2 - 4(-u)}}{2} $$ In this case, we'll use the positive square root since x can't be negative. Thus, $$ x = \frac{-1 + \sqrt{1+4u}}{2} $$ Next, we need to find an expression for \(dx\), using the derivative of \(u\). $$ \frac{du}{dx} = 2x + 1 \\ dx = \frac{du}{2x + 1} $$
03

Substitute for x and dx

Let us substitute our expressions for \(x\) and \(dx\) in the integral. $$ \int \frac{x^{3}+1}{\sqrt{u}} \frac{du}{2x + 1} $$ Substitute x expression: $$ \int \frac{(\frac{-1 + \sqrt{1+4u}}{2})^3+1}{\sqrt{u}} \frac{du}{2(\frac{-1 + \sqrt{1+4u}}{2}) + 1} $$
04

Simplify the integrand and integrate

Now, we simplify the integral expression: $$ \int \frac{((1+4u)^\frac{3}{2} - 3(1+4u)^\frac{1}{2} + 3 )}{2\sqrt{u}} \frac{du}{4u+1} $$ The \(4u+1\) denominator cancels out, leaving the following simplified expression: $$ \frac{1}{2}\int ((1+4u)^\frac{3}{2} - 3(1+4u)^\frac{1}{2} + 3 ) u^{-\frac{1}{2}} du $$ We can now find the antiderivative: $$ \frac{1}{2}\left(\frac{2}{5}(1+4u)^\frac{5}{2} - 2(1+4u)^\frac{3}{2} + 6\sqrt{u}\right) + C $$
05

Substitute u back for x

Finally, replace u in the antiderivative with the original expression in terms of x: $$ \frac{1}{2} \left(\frac{2}{5}(x^2+x)^\frac{5}{2} - 2(x^2+x)^\frac{3}{2} + 6\sqrt{x^2+x}\right) + C $$ Now we have the solution for the given integral: $$ \frac{1}{5}(x^2+x)^\frac{5}{2} - (x^2+x)^\frac{3}{2} + 3\sqrt{x^2+x} + C $$

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x\) (ii) \(\int \frac{d x}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x\) (iii) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (iv) \(\int\left(\frac{1-x^{-2}}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}+\frac{x^{-2}-x}{x^{1 / 2}-x^{-1 / 2}}\right) d x\)

Evaluate the following integrals: (i) \(\int \frac{d x}{(1+x)^{3 / 2}+(1+x)^{1 / 2}}\) (ii) \(\int \frac{\mathrm{dx}}{\sqrt[4]{5-x}+\sqrt{5-x}}\) (iii) \(\int \frac{\mathrm{dx}}{\sqrt{(\mathrm{x}+2)}+\sqrt[4]{(\mathrm{x}+2)}}\) (iv) \(\int \frac{\sqrt{x+1}+2}{(x+1)^{2}-\sqrt{x+1}} d x\)

Evaluate the following integrals: (i) \(\int \frac{x d x}{\left(x^{2}-3 x+2\right) \sqrt{x^{2}-4 x+3}}\) (ii) \(\int \frac{\left(x^{2}+1\right) d x}{\left(x^{2}+2 x+2\right) \sqrt{(x+1)}}\) (iii) \(\int \frac{(2 x+3) d x}{\left(x^{2}+2 x+3\right) \sqrt{x^{2}+2 x+4}}\)

Two of these antiderivatives are elementary functions; find them. (A) \(\int \ln x d x\) (B) \(\int \frac{\ln x d x}{x}\) (C) \(\int \frac{d x}{\ln x}\)

Applying Ostrogradsky's method, find the following integrals: (i) \(\int \frac{d x}{(x+1)^{2}\left(x^{2}+1\right)^{2}}\) (ii) \(\int \frac{d x}{\left(x^{4}+1\right)^{2}}\) (iii) \(\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+1\right)^{4}}\) (iv) \(\int \frac{x^{4}-2 x^{2}+2}{\left(x^{2}-2 x+2\right)^{2}} d x\)
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