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1. Homework Statement :
using the substitution [itex]u=3x+4[/itex], work out:
[itex]\int 2x \sqrt{3x+4}[/itex]2. The attempt at a solution:
[itex]\int 2x \sqrt{3x+4}[/itex]
[itex]u=3x+4 \rightarrow \mathrm{d}x = \frac{\mathrm{d}u}{3}[/itex]
[itex]\int \left( \frac{u-4}{3} \right)\left(u^{\frac12}\right) \ \frac{\mathrm{d}u}{3}[/itex]
[itex]\frac19 \int u^{\frac32} - 4u^{\frac12} \ \mathrm{d}u[/itex]
[itex]=\frac19 \left[ \frac{2u^{\frac52}}{5} - \frac{8u^{\frac32}}{3}\right] + c[/itex]
[itex]=\frac{2u^{\frac52}}{45} - \frac{8u^{\frac32}}{27}}+c[/itex]
[itex]=\frac{2(3x+4)^{\frac52}}{45} - \frac{8(3x+4)^{\frac32}}{27} + c[/itex]3. The problem that I am encountering:
I was close to the answer but it is incorrect. The correct answer is:
[itex]\frac{4(3x+4)^{\frac52}}{45} - \frac{16(3x+4)^{\frac32}}{27} + c[/itex]
Where have I gone wrong? Thanks in advance.
using the substitution [itex]u=3x+4[/itex], work out:
[itex]\int 2x \sqrt{3x+4}[/itex]2. The attempt at a solution:
[itex]\int 2x \sqrt{3x+4}[/itex]
[itex]u=3x+4 \rightarrow \mathrm{d}x = \frac{\mathrm{d}u}{3}[/itex]
[itex]\int \left( \frac{u-4}{3} \right)\left(u^{\frac12}\right) \ \frac{\mathrm{d}u}{3}[/itex]
[itex]\frac19 \int u^{\frac32} - 4u^{\frac12} \ \mathrm{d}u[/itex]
[itex]=\frac19 \left[ \frac{2u^{\frac52}}{5} - \frac{8u^{\frac32}}{3}\right] + c[/itex]
[itex]=\frac{2u^{\frac52}}{45} - \frac{8u^{\frac32}}{27}}+c[/itex]
[itex]=\frac{2(3x+4)^{\frac52}}{45} - \frac{8(3x+4)^{\frac32}}{27} + c[/itex]3. The problem that I am encountering:
I was close to the answer but it is incorrect. The correct answer is:
[itex]\frac{4(3x+4)^{\frac52}}{45} - \frac{16(3x+4)^{\frac32}}{27} + c[/itex]
Where have I gone wrong? Thanks in advance.