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!