Integral x^2 * sqrt(x^3 + 5) dx, using u = x^3 + 5

[asnxbbyx113]

Evaluate the integral by using u=x^3+5.

Integral x^2*sqrt(x^3+5) dx

I did it, and my answer is 2/3(x^3+5)^(3/2)+C, but the answer is wrong. I found du=x^2 dx. Then I got integral sqrt(u) du, and I plugged u in.

 

[soroban]

Hello, asnxbbyx113!

Evaluate the integral by using $\displaystyle u\:=\:x^3\,+\,5.$

. . \(\displaystyle \L\int\)$\displaystyle x^2\sqrt{x^3\,+\,5}\,dx$

I did it, and my answer is: $\displaystyle \,\frac{2}{3}(x^3\,+\,5)^{\frac{3}{2}}\,+\,CC$, but the answer is wrong.

I found: $\displaystyle \,du\:=\:x^2\,dx\;\;$ . . . There's your error!

If $\displaystyle u \:=\:x^3\,+\,5$, then $\displaystyle du \:=\:$3$\displaystyle x^2\,dx$

 

[asnxbbyx113]

so would my answer be 2(x^3+5)^(3/2)+C?

 

[tkhunny]

You are going to have to tell us what your answer is. How did you get that? Was your process sound? Did you make any errors?

 

[morson]

so would my answer be 2(x^3+5)^(3/2)+C?

Check by differentiating your answer!