Ten direct integrals and reverse chain rules

Expand the integrand as $x^{3/2}+3x^{1/2}$ and integrate each power: $x^{5/2}/(5/2)+3x^{3/2}/(3/2)$. The square root restricts the original real input to $x\geq0$.

The integral is $\dfrac25x^{5/2}+2x^{3/2}+C$.

The square expands to $y+2+1/y$. Integrate term by term. The original reciprocal square root requires $y>0$, so the printed $\ln|y|$ is also $\ln y$ on that domain.

The integral is $\dfrac{y^2}2+2y+\ln|y|+C$.

Insert the inner derivative: $\dfrac12\int2(x+1)(x^2+2x+3)^{-3}\,dx$. The reverse chain rule gives $\dfrac1{2(-2)}(x^2+2x+3)^{-2}$.

The integral is $-\dfrac1{4(x^2+2x+3)^2}+C$.

View this as $\int(1/x)(\ln x)\,dx$ and use the derivative of ln x. The real input requires x>0.

The integral is $\dfrac{(\ln x)^2}2+C$.

Rewrite $3t/(1-t)=3(-1+1/(1-t))$. Integrate the constant and account for the inner derivative −1 in the logarithmic term; t≠1.

The integral is $-3t-3\ln|1-t|+C$.

Rewrite the numerator as $(1/x)\,dx$, the differential of ln x. Apply the logarithmic rule to ln x; the original domain is x>0, x≠1.

The integral is $\ln|\ln x|+C$.

The numerator differentiates the denominator. The source first uses $\ln|1+e^x|$ and then removes the absolute value because eˣ>0 for every real x.

The integral is $\ln(1+e^x)+C$.

Write the integral as $2\int\dfrac1{2\sqrt x}e^{\sqrt x}\,dx$. The reciprocal factor is the derivative of √x; x>0.

The integral is $2e^{\sqrt x}+C$.

Write $-\int(-1/x^2)e^{1/x}\,dx$ and use the derivative of 1/x. Retain x≠0.

The integral is $-e^{1/x}+C$.

For base 2017 the derivative includes both ln2017 and the inner derivative 2017. Divide by both factors.

The integral is $\dfrac{2017^{2017x}}{2017\ln2017}+C$.