What is the integral of (7x^2+16)/(x^4+4x^2) ?

The integral of (7x^2+16)/(x^4+4x^2) is -4/x+3/2 arctan(x/2)+C

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integral of (7x^2+16)/(x^4+4x^2)

\int \frac{7 x ^{2} + 16}{x ^{4} + 4 x ^{2}} d x

- \frac{4}{x} + \frac{3}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{7 x ^{2} + 16}{x ^{4} + 4 x ^{2}} d x

Apply Integral Substitution

= \int \frac{7 7 ( u ^{2} + 1 )}{u ^{2} ( 4 u ^{2} + 7 )} d u

Take the constant out:

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

= 77 \cdot \int \frac{u ^{2} + 1}{u ^{2} ( 4 u ^{2} + 7 )} d u

Take the partial fraction of

\frac{u ^{2} + 1}{u ^{2} ( 4 u ^{2} + 7 )} : \frac{1}{7 u ^{2}} + \frac{3}{7 ( 4 u ^{2} + 7 )}

= 77 \cdot \int \frac{1}{7 u ^{2}} + \frac{3}{7 ( 4 u ^{2} + 7 )} d u

Apply the Sum Rule:

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

= 77 (\int \frac{1}{7 u ^{2}} d u + \int \frac{3}{7 ( 4 u ^{2} + 7 )} d u)

\int \frac{1}{7 u ^{2}} d u = - \frac{1}{7 u}

\int \frac{3}{7 ( 4 u ^{2} + 7 )} d u = \frac{3}{14 7} arctan (\frac{2}{7} u)

= 77 (- \frac{1}{7 u} + \frac{3}{14 7} arctan (\frac{2}{7} u))

Substitute back

u = \frac{7}{4} x

= 77 (- \frac{1}{7 \cdot \frac{7}{4} x} + \frac{3}{14 7} arctan (\frac{2}{7} \cdot \frac{7}{4} x))

Simplify

77 (- \frac{1}{7 \cdot \frac{7}{4} x} + \frac{3}{14 7} arctan (\frac{2}{7} \cdot \frac{7}{4} x)) : - \frac{4}{x} + \frac{3}{2} arctan (\frac{x}{2})

= - \frac{4}{x} + \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{4}{x} + \frac{3}{2} arctan (\frac{x}{2}) + C

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What is the integral of (7x^2+16)/(x^4+4x^2) ?
The integral of (7x^2+16)/(x^4+4x^2) is -4/x+3/2 arctan(x/2)+C