Find the integral of 5y ln y dy. | s(y ln y) dy +C

Find the integral of 5y ln y dy. | s(y ln y) dy +C

The Correct Answer and Explanation is:

To evaluate the integral of 5yln⁡y dy5y \ln y \, dy5ylnydy, we will use integration by parts, which is based on the formula:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu

Step 1: Set up uuu and dvdvdv

We choose u=ln⁡yu = \ln yu=lny, because its derivative simplifies the expression, and dv=5y dydv = 5y \, dydv=5ydy, because the integral of yyy is straightforward.

• u=ln⁡yu = \ln yu=lny so that du=1y dydu = \frac{1}{y} \, dydu=y1​dy
• dv=5y dydv = 5y \, dydv=5ydy so that v=5y22v = \frac{5y^2}{2}v=25y2​

Step 2: Apply the Integration by Parts Formula

Substitute into the formula for integration by parts:∫5yln⁡y dy=5y22ln⁡y−∫5y22⋅1y dy\int 5y \ln y \, dy = \frac{5y^2}{2} \ln y – \int \frac{5y^2}{2} \cdot \frac{1}{y} \, dy∫5ylnydy=25y2​lny−∫25y2​⋅y1​dy

Simplifying the second integral:∫5y22⋅1y dy=∫5y2 dy\int \frac{5y^2}{2} \cdot \frac{1}{y} \, dy = \int \frac{5y}{2} \, dy∫25y2​⋅y1​dy=∫25y​dy

Now integrate 5y2\frac{5y}{2}25y​:∫5y2 dy=5y24\int \frac{5y}{2} \, dy = \frac{5y^2}{4}∫25y​dy=45y2​

Step 3: Put Everything Together

Now, substitute this back into the original expression:∫5yln⁡y dy=5y22ln⁡y−5y24+C\int 5y \ln y \, dy = \frac{5y^2}{2} \ln y – \frac{5y^2}{4} + C∫5ylnydy=25y2​lny−45y2​+C

This is the final result for the integral of 5yln⁡y dy5y \ln y \, dy5ylnydy.

Conclusion:

The integral of 5yln⁡y dy5y \ln y \, dy5ylnydy is:5y22ln⁡y−5y24+C\frac{5y^2}{2} \ln y – \frac{5y^2}{4} + C25y2​lny−45y2​+C

where CCC is the constant of integration. This solution combines integration by parts and simplifies the terms to give the final expression.