we have the expression

using a trigonometric substitution
Let

substitute in the original expression

Remember that







Remember that


using the triangle
Find out the value of H
Applying the Pythagorean Theorem
H^2=(3x)^2+1^2
H^2=9x^2+1
H=√(9x^2+1)
![\begin{gathered} \sin u=\frac{3x}{\sqrt[]{9x^2+1}} \\ \sec u=\sqrt[]{9x^2+1} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dq6baqqtu1dx5srbk5vtpyq5i2geicxmcs.png)
substitute
![(1)/(27)\lbrack\ln (\tan u+\sec u)-\sin u\rbrack=(1)/(27)\lbrack\ln (3x+\sqrt[]{9x^2+1})-\frac{3x}{\sqrt[]{9x^2+1}}\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/g30oqks8mtpkm1ktm3706ozd7vqi4jw9si.png)
simplify
![(1)/(27)\lbrack\ln (3x+\sqrt[]{9x^2+1})-\frac{3x}{\sqrt[]{9x^2+1}}\rbrack=\frac{\ln (3x+\sqrt[]{9x^2+1})}{27}-\frac{x}{9\sqrt[]{9x^2+1}}+C](https://img.qammunity.org/2023/formulas/mathematics/college/ixosgjq98mty795ji1cowvyhm4jcxach11.png)
therefore
the answer is
![\frac{\ln(3x+\sqrt[]{9x^2+1})}{27}-\frac{x}{9\sqrt[]{9x^2+1}}+C](https://img.qammunity.org/2023/formulas/mathematics/college/ed2znjuv0rsetg1w0rs2nml132xac31bg8.png)