Integration by substitution is a more powerful integration tool than other people may realize; there are lots of “seemingly unsolvable” integrals that can be done this way. I learned most of these techniques in university, since they provided us with more challenging integrals than usual. The trick is to treat $u$ and $du$ like variables, the to remove all the $x$’s. We’ll start with easier ones:
Take the integral $$\int x\sqrt{x + 4} \, dx.$$ At first, it may seem impossible to use $u$-sub to solve this. If we take $u = x + 4$, then $du = dx$; but $x$ is nowhere to be found in the $du$ expression. However, if we instead use the form $x = u - 4$, then we get: $$\int (u - 4)\sqrt{u} \, du.$$ This, hopefully is easier to solve now. We just use the typical antidifferentiation formulas as follows: $$ \begin{align} \int (u - 4)\sqrt{u} \, du &= \int u^{3/2} - 4u^{1/2} \, du \\ &= \frac{2u^{3/2}}{5} - \frac{8u^{1/2}}{3} + C \\ &= \frac{2(x + 4)^{5/2}}{5} - \frac{8(x +4)^{3/2}}{3} + C. \end{align}$$
This one is similar to the previous, one just slightly more challenging. Try $$\int \frac{x^3}{\sqrt{1 - x^2}} \, dx. $$
Before you read the solution, try experimenting with different values for $u$ and $du$, and play around with the expressions resulting with them. Again, the goal is to come up with a simpler expression to integral by replacing all the terms in $x$ and $dx$ with terms in $u$ and $du$. Did you succeed?
To solve the integral, we take $u = 1 - x^2$ and $du = -2x \, dx$ (that is, $dx = -\frac{1}{2x} \, du$). Thus, $$\begin{align}\int \frac{x^3}{\sqrt{1 - x^2}} \, dx = \int \frac{x^3}{\sqrt{u}} \cdot -\frac{1}{2x}\, du = -\frac{1}{2} \int \frac{x^2}{\sqrt{u}} \, du \end{align}.$$ But we still have not got rid of the $x^2$-term. So we use the fact that $x^2 = 1 - u$. (Why?) Therefore, $$\begin{align} -\frac{1}{2} \int \frac{x^2}{\sqrt{u}} \, du &= -\frac{1}{2} \int \frac{1 - u}{\sqrt{u}} \, du \\ &= -\frac{1}{2} \int u^{-1/2} - u^{1/2} \, du \\ &= -\frac{1}{2} \left(2u^{1/2} - \frac{2u^{3/2}}{3}\right) + C \\ &= -\frac{1}{2} \left[2(1 - x^2)^{1/2} - \frac{2(1 - x^2)^{3/2}}{3}\right] + C. \\ \end{align}.$$
Evaluate the integral $$\int e^\sqrt{x} \, dx.$$ The most natural value of $u$ here is $\sqrt{x}$; thus $du = \frac{1}{2\sqrt{x}} \, dx$. Since $u = \sqrt{x}$, $du = \frac{1}{2u} \, dx$ and finally, $dx = 2u \, du$. Therefore, $$\begin{align} \int e^\sqrt{x} \, dx &= 2\int ue^u du. \end{align}$$ We can simply apply integration by parts here. Thus, $$\begin{align} 2u\int e^u du &= 2\left(ue^u - \int e^u\, du \right) \\ &= 2(ue^u - e^u) + C \\ &= 2(e^\sqrt{x}\sqrt{x} - e^\sqrt{x}) + C. \end{align}$$
$$\int \frac{2x^2}{x^2 + 1} \, dx$$
will be continued... maybe you are able to solve this while i still haven't written the solution
