To evaluate the integral \( \int \frac{x^3 \tan^{-1}(x^4)}{1 + x^8} \,dx, \) we will use integration by parts.
Let's denote:
Substitute into the integral:
\( v = \frac{1}{4} \int \frac{dw}{1+w^2} = \frac{1}{4} \tan^{-1}(w) + C = \frac{1}{4} \tan^{-1}(x^4) + C. \)
Now apply integration by parts:
\( \int u \, dv = uv - \int v \, du \)
Since \( \int v \, du = \int \frac{x^3 \tan^{-1}(x^4)}{1+x^8} dx, \) this integral was our original integral. Thus the integral evaluates to:
\( I = \frac{1}{2} \times \frac{1}{4} (\tan^{-1}(x^4))^2 + C = \frac{1}{8} (\tan^{-1}(x^4))^2 + C. \)
Thus, the result is:
\( \boxed{\frac{(\tan^{-1}(x^4))^2}{8} + c} \)
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
An inductor and a resistor are connected in series to an AC source of voltage \( 144\sin(100\pi t + \frac{\pi}{2}) \) volts. If the current in the circuit is \( 6\sin(100\pi t + \frac{\pi}{2}) \) amperes, then the resistance of the resistor is: