Integral Arctan Sqrtx 1671D2

1. **State the problem:** We need to evaluate the integral $$\int \frac{\arctan(\sqrt{x})}{\sqrt{x}} \, dx$$. 2. **Substitution:** Let $$t = \sqrt{x} = x^{1/2}$$, so $$x = t^2$$ and $$dx = 2t \, dt$$. 3. **Rewrite the integral in terms of $$t$$:** $$\int \frac{\arctan(t)}{t} \, dx = \int \frac{\arctan(t)}{t} \cdot 2t \, dt = \int 2 \arctan(t) \, dt$$ 4. **Simplify the integral:** $$\int 2 \arctan(t) \, dt = 2 \int \arctan(t) \, dt$$ 5. **Use integration by parts:** Let $$u = \arctan(t)$$ and $$dv = dt$$. Then $$du = \frac{1}{1+t^2} dt$$ and $$v = t$$. 6. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int \arctan(t) \, dt = t \arctan(t) - \int \frac{t}{1+t^2} dt$$ 7. **Evaluate the remaining integral:** $$\int \frac{t}{1+t^2} dt$$ Use substitution $$w = 1 + t^2$$, so $$dw = 2t dt$$, thus $$t dt = \frac{dw}{2}$$. Therefore, $$\int \frac{t}{1+t^2} dt = \int \frac{1}{w} \cdot \frac{dw}{2} = \frac{1}{2} \int \frac{1}{w} dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+t^2) + C$$ 8. **Substitute back:** $$\int \arctan(t) \, dt = t \arctan(t) - \frac{1}{2} \ln(1+t^2) + C$$ 9. **Multiply by 2:** $$2 \int \arctan(t) \, dt = 2t \arctan(t) - \ln(1+t^2) + C$$ 10. **Replace $$t = \sqrt{x}$$:** $$\int \frac{\arctan(\sqrt{x})}{\sqrt{x}} \, dx = 2 \sqrt{x} \arctan(\sqrt{x}) - \ln(1+x) + C$$ **Final answer:** $$\boxed{\int \frac{\arctan(\sqrt{x})}{\sqrt{x}} \, dx = 2 \sqrt{x} \arctan(\sqrt{x}) - \ln(1+x) + C}$$