Integration II

Divide by the coefficient of $x$ in the exponent. Check by differentiating: $\dfrac{4e^{4x}}{4} = e^{4x}$ ✓

$[\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1$.

The numerator is the derivative of the denominator, so this has the form $\int \frac{f'}{f}$. No modulus is needed, since $x^2+1 > 0$.

With $u = x^2+1$, $du = 2x\,dx$, the integral becomes $\int u^5\,du$.

By parts with $u = x$ and $\frac{dv}{dx} = \sin x$, so $v = -\cos x$: $-x\cos x + \int\cos x\,dx$.

From $1 \equiv A(x+2) + B(x-1)$: at $x=1$, $A = \tfrac13$; at $x=-2$, $B = -\tfrac13$.