First, note that the function $\cos (x^2 + 1)$ is a composition. The inside function is $x^2 + 1$. Moreover, the derivative of the inside function is $2x$ which is also present in our integrand. These observations suggest that we should try the substitution $u = x^2 + 1$. If we let $u = x^2 + 1$, then using the differential form of the derivative, we have $$du = 2x \ dx$$

Now, the original integral which was in terms of the variable $x$ and the differential $dx$ can be rewritten in terms of the new variable $u$ and its differential $du$. Substituting $u$ for $x^2 + 1$ and $du$ for $2x \ dx$, the original integral becomes $$\int 2x\cos (x^2 + 1) \ dx = \int \cos (x^2 + 1) \ 2x\ dx = \int \cos u \ du.$$ We now have a basic integral in the variable $u$ which can easily be computed: $$\int \cos u \ du = \sin u + C,$$ This is essentially our answer except that it is in the wrong variable. We must go back to $x$ using $u = x^2 + 1$: $$\sin u + C = \sin (x^2 + 1) + C.$$ Thus, using $u$-substitution we can conclude that $$\int 2x\cos (x^2 + 1) \ dx = \sin (x^2 + 1) + C.$$ As a final remark, we can check our answer by differentiation: $$\frac {d}{dx} \sin (x^2 + 1) = \cos (x^2 +1) \cdot 2x \;\; \text {by the chain rule}$$ This agrees with our original integrand, so our answer is correct!

and the integral can be written as $$\int 2x(x^2 + 1)^4 \ dx = \int (x^2 + 1)^4 \ 2x \ dx = \int u^4 \ du.$$ The last integral can be computed as $$\int u^4 \ du = \tfrac {u^5}{5} + C,$$ and by back substituting, we have $$\tfrac {u^5}{5} + C = \tfrac 15(x^2 + 1)^5 + C .$$ Thus, using $u$-substitution we can conclude that $$\int 2x(x^2 + 1)^4 \ dx = \tfrac 15(x^2 + 1)^5 + C.$$

Thus, using $u$-substitution we can conclude that $$\int _0^2 2xe^{x^2 + 1} \ dx = e^5 - e.$$

$$\int 2x\sqrt {x^2 + 1} \ dx = \int \sqrt {x^2 + 1} \cdot 2x\ dx = \int \sqrt {u} \ du.$$ The last integral can be computed as $$\int \sqrt u \ du = \int u^{1/2} \ du = \tfrac {u^{3/2}}{3/2} + C = \tfrac 23 u^{3/2} + C,$$ and by back substituting, we have $$\tfrac 23 u^{3/2} + C = \frac 23 (x^2 + 1)^{3/2} + C.$$ Thus, using $u$-substitution we can conclude that $$\int 2x\sqrt {x^2 + 1} \ dx = \tfrac 23 (x^2 + 1)^{3/2} + C.$$

$$\int \frac {2x}{x^2 + 1} \ dx = \int \frac {1}{x^2 + 1} \cdot 2x\ dx = \int \frac {1}{u} \ du.$$ The last integral can be computed as $$\int \frac {1}{u} \ du = \ln |u| + C,$$ and by back substituting, we have $$\ln |u| + C = \ln |x^2 + 1| + C=\ln (x^2 + 1) + C.$$ Thus, using $u$-substitution we can conclude that $$\int \frac {2x}{x^2 + 1} \ dx = \ln (x^2 + 1) + C.$$

and the integral can be written as $$\int e^x\sin (e^x) \ dx = \int \sin (e^x) \cdot e^x \ dx = \int \sin (u) \ du.$$ The last integral can be computed as $$\int \sin u \ du = -\cos u + C,$$ and by back substituting, we have $$-\cos u + C = -\cos (e^x) + C.$$ Thus, using $u$-substitution we can conclude that $$\int e^x\sin (e^x) \ dx = -\cos (e^x) + C.$$

and the integral can be written as \begin{align*} \int 3x^2\sec (x^3)\tan (x^3) \ dx &= \int \sec (x^3)\tan (x^3) \cdot 3x^2 \ dx\\ &= \int \sec u\tan u \ du. \end{align*}

The last integral can be computed as $$\int \sec u\tan u \ du = \sec u + C,$$ and by back substituting, we have $$\sec u + C = \sec (x^3) + C.$$ Thus, using $u$-substitution we can conclude that $$\int 3x^2\sec (x^3)\tan (x^3) \ dx = \sec (x^3) + C.$$

and the integral can be written as \begin{align*} \int x^3\cos (x^4) \ dx &= \int \cos (x^4) \cdot x^3\ dx \\ &= \int \cos (x^4)\cdot \tfrac 14 \cdot 4x^3\ dx\\ &= \int \tfrac 14\cos u \ du. \end{align*}

The last integral can be computed as $$\int \tfrac 14 \cos u \ du = \tfrac 14 \sin u + C,$$ and by back substituting, we have $$\tfrac 14 \sin u + C = \tfrac 14 \sin (x^4) + C.$$ Thus, using $u$-substitution we can conclude that $$\int x^3\cos (x^4) \ dx = \tfrac 14 \sin (x^4) + C.$$

The last integral can be computed as $$\int -\tfrac 12 e^u \ du = -\tfrac 12 e^u + C,$$ and by back substituting, we have $$-\tfrac 12 e^u + C = -\tfrac 12 e^{-x^2} + C.$$ Thus, using $u$-substitution we can conclude that $$\int xe^{-x^2} \ dx = -\tfrac 12 e^{-x^2} + C.$$

Step 2) Anti-differentiate: $$\int \frac {1}{u^2} \, du \, = \, \int u^{-2} \, du\, =\, \frac {u^{-1}}{-1} + C\, =\, -\frac {1}{u} + C$$ Step 3) Back substitute: $$-\frac {1}{u} + C \, =\, -\frac {1}{\ln x} + C$$ Hence, $$\int \frac {1}{x\ln ^2 x} \ dx \,= \,-\frac {1}{\ln x} + C$$

Let $u = \ln x$. Then we have, $$u = \ln x$$ $$du = \tfrac 1x \ dx$$ and the integral can be written as $$\int \frac {1}{x\ln ^2 x} \ dx = \int \frac {1}{\ln ^2 x} \cdot \frac {1}{x}\ dx = \int \frac {1}{u^2} \ du.$$ The last integral can be computed as $$\int \frac {1}{u^2} \ du = \int u^{-2} \ du = \tfrac {u^{-1}}{-1} + C = -\frac {1}{u} + C,$$ and by back-substituting, we have $$-\frac {1}{u} + C = -\frac {1}{\ln x} + C.$$ Thus, using $u$-substitution we can conclude that $$\int \frac {1}{x\ln ^2 x} \ dx = -\frac {1}{\ln x} + C.$$

The last integral can be computed as $$-\int \frac {1}{u} \ du = -\ln |u| + C,$$ and by back-substituting, we have $$-\ln |u| + C = -\ln |\cos x| + C = \ln |\sec x| +C.$$ Thus, using $u$-substitution we can conclude that $$\int \tan x \ dx = \ln |\sec x| + C.$$