What is the integral of (4(x^2+1))/(x^2) ?

The integral of (4(x^2+1))/(x^2) is 4(-1/x+x)+C

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integral of (4(x^2+1))/(x^2)

\int \frac{4 ( x ^{2} + 1 )}{x ^{2}} d x

4 (- \frac{1}{x} + x) + C

Solution steps

\int \frac{4 ( x ^{2} + 1 )}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x ^{2} + 1}{x ^{2}} d x

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

= 4 \cdot \int \frac{1}{x ^{2}} + 1 d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

= 4 \cdot \int x^{- 2} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int x^{- 2} d x + \int 1 d x)

\int x^{- 2} d x = - \frac{1}{x}

\int 1 d x = x

= 4 (- \frac{1}{x} + x)

Add a constant to the solution

= 4 (- \frac{1}{x} + x) + C

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What is the integral of (4(x^2+1))/(x^2) ?
The integral of (4(x^2+1))/(x^2) is 4(-1/x+x)+C