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

The integral of (7(x^2+4))/(x(x^2+7)) is 3/2 ln|x^2+7|+4ln|x|+C

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

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

\frac{3}{2} ln x^{2} + 7 + 4 ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

= 7 \cdot \int \frac{x ^{2} + 4}{x ( x ^{2} + 7 )} d x

Expand

\frac{x ^{2} + 4}{x ( x ^{2} + 7 )} : \frac{x ^{2}}{x ( x ^{2} + 7 )} + \frac{4}{x ( x ^{2} + 7 )}

Apply the Sum Rule:

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

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

\int \frac{x ^{2}}{x ( x ^{2} + 7 )} d x = \frac{1}{2} ln x^{2} + 7

\int \frac{4}{x ( x ^{2} + 7 )} d x = 4 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{14} ln x^{2} + 7)

= 7 (\frac{1}{2} ln x^{2} + 7 + 4 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{14} ln x^{2} + 7))

Simplify

7 (\frac{1}{2} ln x^{2} + 7 + 4 (\frac{1}{7} ln ∣ x ∣ - \frac{1}{14} ln x^{2} + 7)) : \frac{3}{2} ln x^{2} + 7 + 4 ln ∣ x ∣

= \frac{3}{2} ln x^{2} + 7 + 4 ln ∣ x ∣

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 7 + 4 ln ∣ x ∣ + C

What is the integral of (7(x^2+4))/(x(x^2+7)) ?
The integral of (7(x^2+4))/(x(x^2+7)) is 3/2 ln|x^2+7|+4ln|x|+C