Having trouble with Integration by Parts

[relinquished™]

I'm real stuck with this problem of mine in The Calculus 7 by Leithold


\int arctan \sqrt{x} dx


Since there is no elementary formula for integration of an inverse trigo function, we cannot manipulate the integrand in such a way as to integrate easily with one step of Integration by parts.(please verify)

So first, If we let u = arctan \sqrt{x} and
dv = dx

then

du = \frac{dx}{2 \sqrt{x} (x+1)} and
v=x

then we have


\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}


Now, Case 1, without simplifying \frac{x}{\sqrt {x}}

let db = \frac{1}{2 \sqrt{x} (x+1)}dx
and a = x.
Then da = dx
and b = arctan \sqrt{x}

so


\int arctan \sqrt{x} dx = xarctan \sqrt{x} - xarctan \sqrt{x} + \int arctan \sqrt{x} dx

which of course will lead us nowhere but 0=0

Now, Case 2 - If we simplify \frac{x}{\sqrt {x}} = \sqrt{x}

let a= \sqrt{x} and db = \frac{dx}{1+x} ,
then da = \frac{dx}{2 \sqrt{x}} and b = ln (x+1)

so


\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \frac{1}{2} ( \sqrt{x}ln (x+1) - \int \frac{ln |1+x| dx}{2 \sqrt{x}})

and if we do Integration by parts again, we will just get an infinite sequence of \sqrt{x}ln (x+1) (please verify)

I'm really open to suggestions ^_^`

[irony of truth]

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[irony of truth]

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}

From there... add 1 and -1 to your numerator, and disect the expression into two... hope you figure out the next steps. (in your integral of udv)

[relinquished™]

0.0 hey... why didn't I think of that???

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{xdx}{2 \sqrt{x} (x+1)}

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{x + 1 - 1}{2 \sqrt{x} (x+1)}dx

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{(x + 1)}{2 \sqrt{x} (x+1)} - \frac{1}{2 \sqrt{x} (x+1)}) dx

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int ( \frac{1}{2 \sqrt{x}} - \frac{1}{2 \sqrt{x} (x+1)}) dx

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \int \frac{dx}{2 \sqrt{x}} + \int \frac{dx}{2 \sqrt{x} (x+1)}

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{dx}{2 \sqrt{x} (x+1)})

let u^2 = x, then u = \sqrt{x}, since du = \frac{1}{2 \sqrt{x}} it can be integrated directly and it yields arctan...

so

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + \int \frac{du}{(u^2+1)}

\int arctan \sqrt{x} dx = xarctan \sqrt{x} - \sqrt{x} + arctan \sqrt{x} + C

Is this correct? ^_^;;

Man... I really don't like it when the answer is just right in your face and slaps you XD. Gotta remember the +a -a concept ^_^;;; Thanx btw Irony of Truth

[irony of truth]

now that's correct.