# 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 ^_^`

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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.

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