Final answer to the problem
$\frac{\arctan\left(\frac{x}{\sqrt{10^{-262}+2\cdot 10^{-131}x}}\right)}{\sqrt{10^{-262}+2\cdot 10^{-131}x}}+C_0$
Got another answer? Verify it here!
Step-by-step Solution
How should I solve this problem?
- Choose an option
- 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
- Load more...
Can't find a method? Tell us so we can add it.
1
Expand the expression $\left(10^{-131}+x\right)^2$ using the square of a binomial: $(a+b)^2=a^2+2ab+b^2$
$\int\frac{1}{10^{-262}+2\cdot 10^{-131}x+x^{2}}dx$
Learn how to solve problems step by step online.
$\int\frac{1}{10^{-262}+2\cdot 10^{-131}x+x^{2}}dx$
Learn how to solve problems step by step online. Find the integral int(1/((10^(-131)+x)^2))dx. Expand the expression \left(10^{-131}+x\right)^2 using the square of a binomial: (a+b)^2=a^2+2ab+b^2. 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
$\frac{\arctan\left(\frac{x}{\sqrt{10^{-262}+2\cdot 10^{-131}x}}\right)}{\sqrt{10^{-262}+2\cdot 10^{-131}x}}+C_0$
