integrate, assuming ( x > 0 ).\n int sqrt { x } ln ( 5 x ) d x \n int sqrt { x } ln ( 5 x ) d x = square…

integrate, assuming ( x > 0 ).\n int sqrt { x } ln ( 5 x ) d x \n int sqrt { x } ln ( 5 x ) d x = square \n(type an exact answer. use parentheses to clearly denote the argument of each function.)

integrate, assuming ( x > 0 ).\n int sqrt { x } ln ( 5 x ) d x \n int sqrt { x } ln ( 5 x ) d x = square \n(type an exact answer. use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Apply integration by parts

Integration by parts formula is $\int u\mathrm{d}v=uv - \int v\mathrm{d}u$. Let $u = \ln(5x)$ and $\mathrm{d}v=\sqrt{x}\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v=\int\sqrt{x}\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}$.

Step2: Substitute into integration by parts formula

[ \begin{align*} \int\sqrt{x}\ln(5x)\mathrm{d}x&=\frac{2}{3}x^{\frac{3}{2}}\ln(5x)-\int\frac{2}{3}x^{\frac{3}{2}}\cdot\frac{1}{x}\mathrm{d}x\ &=\frac{2}{3}x^{\frac{3}{2}}\ln(5x)-\frac{2}{3}\int x^{\frac{1}{2}}\mathrm{d}x \end{align*} ]

Step3: Integrate the remaining integral

$\int x^{\frac{1}{2}}\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}+C$. So, [ \begin{align*} \int\sqrt{x}\ln(5x)\mathrm{d}x&=\frac{2}{3}x^{\frac{3}{2}}\ln(5x)-\frac{2}{3}\cdot\frac{2}{3}x^{\frac{3}{2}}+C\ &=\frac{2}{3}x^{\frac{3}{2}}\ln(5x)-\frac{4}{9}x^{\frac{3}{2}}+C \end{align*} ]

Answer:

$\frac{2}{3}x^{\frac{3}{2}}\ln(5x)-\frac{4}{9}x^{\frac{3}{2}}+C$