evaluate the following integral. ∫(10x + 8)ln(7x)dx check answer

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \ln(7x)$ and $dv=(10x + 8)dx$. Then $du=\frac{1}{x}dx$ and $v=\int(10x + 8)dx=5x^{2}+8x$.

Step2: Apply the formula

$\int(10x + 8)\ln(7x)dx=(5x^{2}+8x)\ln(7x)-\int(5x^{2}+8x)\frac{1}{x}dx$.

Step3: Simplify the second integral

$\int(5x^{2}+8x)\frac{1}{x}dx=\int(5x + 8)dx$.

Step4: Integrate the simplified integral

$\int(5x + 8)dx=\frac{5}{2}x^{2}+8x+C$.

Step5: Write the final result

$\int(10x + 8)\ln(7x)dx=(5x^{2}+8x)\ln(7x)-\left(\frac{5}{2}x^{2}+8x\right)+C=(5x^{2}+8x)\ln(7x)-\frac{5}{2}x^{2}-8x + C$.

Answer:

$(5x^{2}+8x)\ln(7x)-\frac{5}{2}x^{2}-8x + C$