Expand the power.

$$\int_{ }^{ }\left(x+1\right)^2dx=\int_{ }^{ }???\ dx$$

Evaluate the integral.

$$\int_{ }^{ }\left(x+1\right)^2dx=\int_{ }^{ }x^2+2x+1\ dx=???$$

Substitute $$t=x+1$$   to integrate $$\int_{ }^{ }\left(x+1\right)^7dx$$   .
Since $$t=x+1$$   , then $$\left(x+1\right)^7=???$$   .

Substitute $$t=x+1$$   to integrate $$\int_{ }^{ }\left(x+1\right)^7dx$$   .
Since $$t=x+1$$   , then $$\left(x+1\right)^7=t^7$$   .

So, $$\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }???dx$$

Since $$t=x+1$$   , when we differentiate we have

$$\frac{dt}{dx}=???$$

Since $$t=x+1$$   , then $$\left(x+1\right)^7=t^7$$   .
and also $$\frac{dt}{dx}=1$$   , so $$dx=dt$$   .

So, $$\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }t^7dt=???$$ .

Since $$t=x+1$$   , then $$\left(x+1\right)^7=t^7$$   .
and also $$\frac{dt}{dx}=1$$   , so $$dx=dt$$   .

So, $$\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }t^7dt=\frac{1}{8}t^8+c=???$$ .

(the original question is in variable "x", so the answer should also be in "x")

Since $$t=x^2-3$$ , then:

$$\frac{dt}{dx}=???$$

Since $$t=x^2-3$$ , then $$\frac{dt}{dx}=2x$$ .

Thus,

$$dx=???\ dt$$

Since $$t=x^2-3$$ , then $$\frac{dt}{dx}=2x$$ .

Thus, $$dx=\frac{1}{2x}\ dt$$ .

So, the integral become:

$$\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{???}}dx$$

Since $$t=x^2-3$$ , then $$\frac{dt}{dx}=2x$$ .

Thus, $$dx=\frac{1}{2x}\ dt$$ .

So, the integral become:

$$\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt=\int_{ }^{ }???dt$$

Since $$t=x^2-3$$ , then $$\frac{dt}{dx}=2x$$ .

Thus, $$dx=\frac{1}{2x}\ dt$$ .

So, the integral become:

$$\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt=\int_{ }^{ }\frac{1\ }{2\sqrt[]{t}}dt=\int_{ }^{ }\frac{1\ }{2}t^{-\frac{1}{2}}dt=???$$

Since $$t=x^2-3$$ , then $$\frac{dt}{dx}=2x$$ .

Thus, $$dx=\frac{1}{2x}\ dt$$ .

So, the integral become:

$$\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt=$$ $$\int_{ }^{ }\frac{1\ }{2\sqrt[]{t}}dt=\int_{ }^{ }\frac{1\ }{2}t^{-\frac{1}{2}}dt=t^{\frac{1}{2}}+c=???$$