Immediate and almost immediate integrals

By linearity the integral of the sum is the sum of the integrals and multiplicative constants come out of the integral sign. From the table of elementary integrals, the integral of $x^3$ is $\frac{x^4}{4}$ and the integral of $x$ is $\frac{x^2}{2}$, therefore the first two terms give $x^4$ and $-x^2$. The integral of $\frac{1}{x}$ is $\ln |x|$, hence the third term gives $3\ln |x|$. The three constants of integration are combined into the single constant $c$. Differentiating $x^4-x^2+3\ln |x|$ gives back the integrand, therefore the result is correct.

The derivative of the exponent $x^2$ is $2x$ and the factor $x$ appears in the integrand, therefore the integral is almost immediate of the form $\int f^{\prime }{e}^{f}dx$ with $f(x)=x^2$. One multiplies and divides by $2$ to reconstruct exactly the factor $f^{\prime }(x)=2x$; at that point the formula gives $e^{x^2}$, multiplied by the coefficient $\frac{1}{2}$ left outside. The check by differentiation gives $\frac{1}{2}{e}^{x^2}\cdot 2x=x{e}^{x^2}$, which is the integrand.

Since $e^{2x}=(e^x{)}^2$, the denominator can be written as $1+(e^x{)}^2$. The numerator $e^x$ is the derivative of $e^x$, therefore the integral has the form $\int \frac{f^{\prime }}{1+f^2}dx$ with $f(x)=e^x$, whose result is $\arctan f(x)$. Differentiating $\arctan (e^x)$ gives $\frac{e^x}{1+e^{2x}}$, hence the result is correct.