What is the integral of sqrt(9x)(1/x+x^2) ?

The integral of sqrt(9x)(1/x+x^2) is 3(2/7 x^3sqrt(x)+2sqrt(x))+C

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integral of sqrt(9x)(1/x+x^2)

\int 9 x (\frac{1}{x} + x^{2}) d x

3 (\frac{2}{7} x^{3} x + 2 x) + C

Solution steps

\int 9 x (\frac{1}{x} + x^{2}) d x

9 x = 3 x

= \int 3 x (\frac{1}{x} + x^{2}) d x

Take the constant out:

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

= 3 \cdot \int x (\frac{1}{x} + x^{2}) d x

Apply Integration By Parts

: x + x^{3} x - \int \frac{5 x ^{3} - 1}{2 x} d x

= 3 (x + x^{3} x - \int \frac{5 x ^{3} - 1}{2 x} d x)

\int \frac{5 x ^{3} - 1}{2 x} d x = \frac{1}{7} x (5 x^{3} - 7)

= 3 (x + x^{3} x - \frac{1}{7} x (5 x^{3} - 7))

Expand

x + x^{3} x - \frac{1}{7} x (5 x^{3} - 7) : \frac{2}{7} x^{3} x + 2 x

= 3 (\frac{2}{7} x^{3} x + 2 x)

Add a constant to the solution

= 3 (\frac{2}{7} x^{3} x + 2 x) + C

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What is the integral of sqrt(9x)(1/x+x^2) ?
The integral of sqrt(9x)(1/x+x^2) is 3(2/7 x^3sqrt(x)+2sqrt(x))+C