I don't even know where to start. Any help on where to start would be greatly appreciated.

If this helps, the solution given by the book is:

(1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c

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I don't even know where to start. Any help on where to start would be greatly appreciated.

If this helps, the solution given by the book is:

(1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c

This is not a partial fraction. Maybe a trig substitution.

\(\displaystyle \int\sqrt{x^{2}+2x}dx=\int\sqrt{x(x+2)}dx\)

This one can be a little tricky though.

Let \(\displaystyle x=u-1, \;\ u+1=x+2, \;\ du=dx\)

Make the subs and we have \(\displaystyle \int\sqrt{(u-1)(u+1)}du=\int\sqrt{u^{2}-1}du\)

Now, it is a form in which we can use trig sub easily enough.

\(\displaystyle \int\sqrt{u^{2}-1}du\)

Let \(\displaystyle u=sec(t), \;\ du=sec(t)tan(t)dt\)

Make the subs:

\(\displaystyle \int\sqrt{sec^{2}(t)-1}sec(t)tan(t)dt\)

But \(\displaystyle sec^{2}(t)-1=tan^{2}(t)\)

\(\displaystyle \int\sqrt{tan^{2}(t)}sec(t)tan(t)dt\)

\(\displaystyle \int tan^{2}(t)sec(t)dt\)

\(\displaystyle \int sec(t)(sec^{2}(t)-1)dt\)

\(\displaystyle \int sec^{3}(t)dt-\int sec(t)dt\)

Now, I will let you finish. OKey-doke?. Use the reduction formula if need be on the sec and you have it. That formula is in any calc book

See?. No partial fractions. When you get that integrated, you will need to resub to get it back in terms of x.

Galactus's solution is good, but if you are not as smart as he is, you can always look atdjo201 said:

I don't even know where to start. Any help on where to start would be greatly appreciated.

If this helps, the solution given by the book is:

(1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c

sqrt(x^2 + 2x)

as an opportunity to complete the square:

sqrt(x^2 + 2x )

sqrt(x^2 + 2x + 1 - 1)

sqrt((x +1)^2 - 1)

Now it becomes obvious to put u = x + 1

Then the trig subst will be obvious, too.

Thanks for the help so far. By reduction formula, do you mean the one from the Table of Integrals. If so, I'm not allowed to use that. Any other way?galactus said:This is not a partial fraction. Maybe a trig substitution.

\(\displaystyle \int\sqrt{x^{2}+2x}dx=\int\sqrt{x(x+2)}dx\)

This one can be a little tricky though.

Let \(\displaystyle x=u-1, \;\ u+1=x+2, \;\ du=dx\)

Make the subs and we have \(\displaystyle \int\sqrt{(u-1)(u+1)}du=\int\sqrt{u^{2}-1}du\)

Now, it is a form in which we can use trig sub easily enough.

\(\displaystyle \int\sqrt{u^{2}-1}du\)

Let \(\displaystyle u=sec(t), \;\ du=sec(t)tan(t)dt\)

Make the subs:

\(\displaystyle \int\sqrt{sec^{2}(t)-1}sec(t)tan(t)dt\)

But \(\displaystyle sec^{2}(t)-1=tan^{2}(t)\)

\(\displaystyle \int\sqrt{tan^{2}(t)}sec(t)tan(t)dt\)

\(\displaystyle \int tan^{2}(t)sec(t)dt\)

\(\displaystyle \int sec(t)(sec^{2}(t)-1)dt\)

\(\displaystyle \int sec^{3}(t)dt-\int sec(t)dt\)

Now, I will let you finish. OKey-doke?. Use the reduction formula if need be on the sec and you have it. That formula is in any calc book

See?. No partial fractions. When you get that integrated, you will need to resub to get it back in terms of x.

Yeah, there is. The old-fasioned way. By hand.

\(\displaystyle \int sec(t)dt=ln(|sec(t)+tan(t)|)\)

Let \(\displaystyle u=sec(t)+tan(t), \;\ du=(sec^{2}(t)+sec(t)tan(t))dt\)

\(\displaystyle \int sec(t)dt=\int sec(t)\left(\frac{sec(t)+tan(t)}{sec(t)+tan(t)}\right)dt\)

\(\displaystyle \int\frac{sec^{2}(t)+sec(t)tan(t)}{sec(t)+tan(t)}dt\)

\(\displaystyle ln|sec(t)+tan(t)|+C\)

You can try the other. It is late and I am going beddy-bye

Thanks for your help.galactus said:Yeah, there is. The old-fasioned way. By hand.

\(\displaystyle \int sec(t)dt=ln(|sec(t)+tan(t)|)\)

Let \(\displaystyle u=sec(t)+tan(t), \;\ du=(sec^{2}(t)+sec(t)tan(t))dt\)

\(\displaystyle \int sec(t)dt=\int sec(t)\left(\frac{sec(t)+tan(t)}{sec(t)+tan(t)}\right)dt\)

\(\displaystyle \int\frac{sec^{2}(t)+sec(t)tan(t)}{sec(t)+tan(t)}dt\)

\(\displaystyle ln|sec(t)+tan(t)|+C\)

You can try the other. It is late and I am going beddy-bye

How is the \(\displaystyle sec^{3}(t)dt\) coming along?.

You can use parts for this one.

Rewrite as:

\(\displaystyle \int (1+tan^{2}(t))sec(t)dt\)

Let \(\displaystyle u=sec(t), \;\ dv=sec^{2}(t)dt, \;\ du=sec(t)tan(t)dt, \;\ v=tan(t)\)

\(\displaystyle \int sec^{3}(t)dt=sec(t)tan(t)-\int sec^{3}(t)dt+\int sec(t)dt\)

Here is something a lot overlook. See your original integral on the right side of the equals?. You can add it to both sides and get:

\(\displaystyle 2\int sec^{3}(t)dt=sec(t)tan(t)+\int sec(t)dt\)

\(\displaystyle \int sec^{3}(t)dt=\frac{1}{2}sec(t)tan(t)+\frac{1}{2}\int sec(t)dt\)

Now it is a matter of going back through and resubbing several times because of the various substitutions.

Thanks. That last part was the one I overlooked.galactus said:How is the \(\displaystyle sec^{3}(t)dt\) coming along?.

You can use parts for this one.

Rewrite as:

\(\displaystyle \int (1+tan^{2}(t))sec(t)dt\)

Let \(\displaystyle u=sec(t), \;\ dv=sec^{2}(t)dt, \;\ du=sec(t)tan(t)dt, \;\ v=tan(t)\)

\(\displaystyle \int sec^{3}(t)dt=sec(t)tan(t)-\int sec^{3}(t)dt+\int sec(t)dt\)

Here is something a lot overlook. See your original integral on the right side of the equals?. You can add it to both sides and get:

\(\displaystyle 2\int sec^{3}(t)dt=sec(t)tan(t)+\int sec(t)dt\)

\(\displaystyle \int sec^{3}(t)dt=\frac{1}{2}sec(t)tan(t)+\frac{1}{2}\int sec(t)dt\)

Now it is a matter of going back through and resubbing several times because of the various substitutions.

After looking at the suggestion PaulK gave, I see you can sub \(\displaystyle x+1=sec(t)\) and get another way of going about it.

That eliminates the radical because \(\displaystyle sec^{2}(t)-1=tan^{2}(t)\)