(√(1+x^2))/x^4 dx is equal to : Option 1: (1+1/x^2)^(3/2)+C :Where C is an arbitrary constant Option 2: -1/3(1+1/x^2)^(3/2)+C :Where C is an arbitrary constant Option 3: 1/3(1+1/x^2)^(3/2)+C :Where C is an arbitrary constant Option 4: -1/3(1+1/x^2)^(5/2)+C :Where C is an arbitrary constant

Answer: -1/3(1+1/x^2)^(3/2)+C :Where C is an arbitrary constant. Solution: Take x out of the root: the integrand is 1/x^3√(1+1/x^2). Put u=1+1/x^2, so du=-2/x^3dx. The integral becomes -1/2√u du=-1/3u^(3/2)+C=-1/3(1+1/x^2)^(3/2)+C.

Calculus

Integrals

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$\int \frac{\sqrt{1+x^2}}{x^4} dx$ is equal to :

$\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{5/2}+C$ :Where C is an arbitrary constant

✅ Correct Option: 2

Take $x$ out of the root: the integrand is $\frac{1}{x^3}\sqrt{1+\frac{1}{x^2}}$ .

Put $u=1+\frac{1}{x^2}$ , so $du=-\frac{2}{x^3}dx$ .

The integral becomes $-\frac{1}{2}\int\sqrt{u}\,du=-\frac{1}{3}u^{3/2}+C=-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$ .

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$\int \frac{\sqrt{1+x^2}}{x^4} dx$ is equal to :