Final answer to the problem
$\frac{\arctan\left(\frac{x}{\sqrt{2+2x}}\right)}{\sqrt{2+2x}}+C_0$
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Step-by-step Solution
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- Integrate by partial fractions
- Integrate by substitution
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Integrate using trigonometric identities
- Integrate using basic integrals
- Product of Binomials with Common Term
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1
Solve the integral by applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$
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$\frac{1}{\sqrt{2+2x}}\arctan\left(\frac{x}{\sqrt{2+2x}}\right)$
Learn how to solve problems step by step online. Find the integral int(1/(2+2xx^2))dx. Solve the integral by applying the formula \displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right). Multiply the fraction by the term . As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C.
Final answer to the problem
$\frac{\arctan\left(\frac{x}{\sqrt{2+2x}}\right)}{\sqrt{2+2x}}+C_0$
