Substitution Techniques for Integration

Some integrals that appear difficult are relatively easy if you use substitution.

Take for example the indefinite integral

$$\int x\sqrt{25+x}dx$$

Let

For integration by parts:

$$u=x,dv=\sqrt{25+x}$$ $$\frac{du}{dx}=1⟹du=dx$$ $$v=\int \sqrt{25+x}dx=\frac{2}{3}(25+x{)}^{2/3}$$

• This gets a bit messy. Let’s try to use substitution instead.

For $u$-sub:

$$u=25+x⟹du=dx$$ $$\int x\sqrt{25+x}dx=\int x\sqrt{u}du$$

• We can’t integrate with respect to $u$ when we have the coefficient $x$. However, notice how we can declare $x$ in terms of $u$.

$$u=25+x⟹x=u-25$$ $$=\int (u-25)\sqrt{u}du=\int u\sqrt{u}du-25\int \sqrt{u}du$$

• This becomes quite trivial to integrate.

$$=\frac{2}{5}{u}^{5/2}+\frac{50}{3}{u}^{3/2}+C$$

• We finish by substituting back in our original variable $x$.

$$=\frac{2}{5}(25+x{)}^{5/2}+\frac{50}{3}(25+x{)}^{3/2}+C$$