Integration by Substitution – Examples with Answers

The success of a change of variable consists in transforming the integral into a simpler integral, such as a power, or the integral of some elementary function. In this example, the following substitution is made:

$latex u =1+2x$

$latex du =2dx$

Therefore:

$latex dx =\dfrac {du}{2}$

The idea is to make the new variable and its derivative appear in the original integrand. Substituting the previous results and taking into account that the numerical coefficients leave out of the integral as they are constant, we obtain:

$$ \int 3(1+2x)^4dx=3 \int u^4\cdot\left(\dfrac {du}{2}\right)=\frac{3}{2} \int u^4du$$

The power rule for integration is then applied:

$$\int kx^ndx=k\left(\dfrac{x^{n+1}}{n+1}\right)+C$$

Where $latex n \neq -1$

In addition, $latex k $ is the coefficient and $latex C$ is the constant of integration.

In this example, $latex n = 4$, therefore:

$$ \int 3(1+2x)^4dx=\frac{3}{2} \int u^4du=\frac{3}{2}\left(\dfrac{u^{4+1}}{4+1}\right)+C$$

$$ \frac{3}{2} \int u^4du=\frac{3}{2}\left(\dfrac{u^5}{5}\right)+C$$

$$ \frac{3}{2} \int u^4du=\frac{3u^5}{10}+C$$

Finally, substitute $latex u =1+2x$ into the result:

$$ \int 3(1+2x)^4dx=\frac{3(1+2x)^5}{10}+C$$

In this type of integral, it gives good results to take the radical quantity as the variable $latex u$, since when deriving, an expression similar to the numerator of the fraction appears:

$latex u =1+x^2$

$latex du =2xdx$

Substituting the proposed change, we obtain:

$$ \int \frac {4x}{\sqrt{1+x^2}}dx=\int \frac {2du}{\sqrt{u}}$$

Now the integrand is written as a power:

$$\int \frac {2du}{\sqrt{u}}=2\int \left(u^{-\frac{1}{2}}\right)du$$

Applying the power rule with $latex n=-\dfrac{1}{2}$:

$$2\int \left(u^{-\frac{1}{2}}\right)du=2\Big[\dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Big]+C=2\Big[\dfrac{u^{\frac{1}{2}}}{\frac{1}{2}}\Big]+C$$

$$2\int \left(u^{-\frac{1}{2}}\right)du=4u^{\frac{1}{2}}+C$$

Finally, we substitute back:

$$ \int \frac {4x}{\sqrt{1+x^2}}dx=4(1+x^2)^{\frac{1}{2}}+C$$

$$ \int \frac {4x}{\sqrt{1+x^2}}dx=4\sqrt{1+x^2}+C$$

The following substitution can be applied:

$latex u = 2x+1$

$latex du =2dx$

This results in:

$latex x =\dfrac{ u-1}{2}$

$latex dx =\dfrac{ du}{2}$

Now the whole integral is transformed to the variable $latex u$:

$$ \int \left[\dfrac{ (u-1)\sqrt{u}}{2}\right]\dfrac{ du}{2}=\frac{1}{4}\int(u-1)\sqrt{u}du$$

$$=\frac{1}{4}\int(u-1)u^\frac{1}{2}du$$

Then the distributive property is applied to the integrand:

$$ \frac{1}{4}\int(u-1)u^\frac{1}{2}du=\frac{1}{4}\int \left(u^\frac{3}{2}-u^\frac{1}{2}\right) du$$

That leads to two simple integrals that are solved with the power rule:

$$ \frac{1}{4}\int \left(u^\frac{3}{2}-u^\frac{1}{2}\right) du=\frac{1}{4}\int u^\frac{3}{2}du-\frac{1}{4}\int u^\frac{1}{2}du$$

$$\frac{1}{4}\int u^\frac{3}{2}du-\frac{1}{4}\int u^\frac{1}{2}du=\frac{1}{4}\left[\dfrac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]-\frac{1}{4}\left[\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+C$$

$$=\frac{1}{4}\left[\dfrac{u^{\frac{5}{2}}}{\frac{5}{2}}\right]-\frac{1}{4}\left[\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]+C$$

$$=\frac{1}{10}u^{\frac{5}{2}}-\frac{1}{6}u^{\frac{3}{2}}+C$$

The result is now returned to the original variable:

$$ \int x\sqrt{2x+1}dx=\frac{1}{10}(2x+1)^{\frac{5}{2}}-\frac{1}{6}(2x+1)^{\frac{3}{2}}+C$$

Optionally, the above expression can be factored:

$$ \int x\sqrt{2x+1}dx=(2x+1)^{\frac{3}{2}}\left[\left(\frac{2x+1}{10}\right)-\frac{1}{6}\right]+C$$

$$ \int x\sqrt{2x+1}dx=\sqrt{(2x+1)^3}\left(\frac{x}{5}-\frac{1}{15}\right)+C$$

$$ \int x\sqrt{2x+1}dx=\sqrt{(2x+1)^3}\left(\frac{3x-1}{15}\right)+C$$

The following change of variable is appropriate since both the natural logarithm and its derivative appear in the integrand:

$latex u =5\ln x+1$

$latex du =\dfrac {5dx}{x}$

Note that:

$latex \dfrac {du}{5}=\dfrac {dx}{x}$

This is substituted into the original integral, which leads to:

$$ \int \frac{(5\ln x+1)^2}{x}dx=\int u^2 \left(\frac {du}{5}\right)$$

$$=\frac{1}{5}\int u^2 du$$

Applying the power rule:

$$ \frac{1}{5}\int u^2 du=\frac{1}{5}\left[\dfrac{u^{2+1}}{2+1}\right]+C=\dfrac{u^3}{15}+C$$

Finally, we substitute back:

$$ \int \frac{(5\ln x+1)^2}{x}dx=\dfrac{(5\ln x+1)^3}{15}+C$$

If the cosine argument is taken as the variable $latex u$, its derivative is present, let’s see:

$latex u = x^4+2$

$latex du =4x^3dx$

$latex \dfrac{du}{4}=x^3dx$

Making the corresponding substitutions in the original integral:

$$ \int x^3\cos(x^4+2)dx=\frac{1}{4}\int \cos u du$$

This is an immediate integral that appears in tables:

$latex \int \cos u du= \sin u + C$

Therefore:

$$\frac{1}{4}\int \cos u du=\frac{1}{4}\sin u + C$$

Returning the change of variables:

$$ \int x^3\cos(x^4+2)dx=\frac{1}{4}\sin (x^4+2)+ C$$

The appropriate variable change is:

$latex u =3x-5$

$latex du =3dx$

Substituting:

$$ \int e^{3x-5}dx=\int e^u\left(\frac{du}{3}\right)$$

$$=\frac{1}{3}\int e^udu$$

This integral is immediate:

$$ \frac{1}{3}\int e^udu=\frac{1}{3}e^u+C$$

Therefore:

$$ \int e^{3x-5}dx=\frac{1}{3}e^{3x-5}+C$$

If we take $latex u$ as the polynomial in the denominator, we immediately see that its derivative appears in the numerator, which leads to a simple integral.

Then, we have:

$latex 2x^2+3x-1$

$latex du = (4x+3)dx$

Therefore:

$$ \int \frac{(4x+3)dx}{2x^2+3x-1}=\int\frac{du}{u}$$

$$=ln\left | 2x^2+3x-1 \right | +C$$

First, the proposed integral is rewritten as follows:

$$\int \frac {dx}{x^2+9}=\int \frac {dx}{9\left(\dfrac{x^2}{9}+1\right)}$$

Where 9 is taken out as the common factor, and it is now easy to see that:

$$\int \frac {dx}{9\left(\dfrac{x^2}{9}+1\right)}=\frac{1}{9}\int \frac {dx}{\left[\left(\dfrac{x}{3}\right)^2+1\right]}$$

$$=\frac{1}{9}\int \dfrac {dx}{\left[1+\left(\dfrac{x}{3}\right)^2\right]}$$

The change of variable is as follows:

$latex u =\dfrac{x}{3}$

$latex du =\dfrac{dx}{3}$

$latex dx =3du$

Thus, the integral becomes:

$$\frac{1}{9}\int \dfrac {dx}{\left[1+\left(\dfrac{x}{3}\right)^2\right]}=\frac{3}{9}\int \dfrac {du}{1+u^2}$$

$$=\frac{1}{3}\int \dfrac {du}{1+u^2}$$

This integral appears in tables:

$$\int \dfrac {dx}{1+x^2}=\arctan x + C$$

Therefore:

$$\int \frac {dx}{x^2+9}=\frac{1}{3}\int \dfrac {du}{1+u^2}$$

$$=\left(\frac{1}{3}\right)\arctan u + C$$

And when substituting back, we get:

$$\int \frac {dx}{x^2+9}=\left(\frac{1}{3}\right)\arctan u + C$$

$$=\left(\frac{1}{3}\right)\arctan\left( \dfrac{x}{3} \right)+ C$$

In general, integrals of the form $latex \int \frac {dx}{a^2+x^2}$, are solved as follows:

$$\int \frac {dx}{a^2+x^2}=\frac{1}{a}\arctan\left( \dfrac{x}{a} \right)+ C$$

Inspired by the previous example, the integral is rewritten as:

$$\int \frac {dx}{4-x^2}=\int \frac {dx}{4\left(1-\dfrac{x^2}{4}\right)}=\frac{1}{4}\int \frac {dx}{\left[1-\left(\dfrac{x}{2}\right)^2\right]}$$

This time, the variable change would be:

$latex u =\dfrac{x}{2}$

$latex du =\dfrac{dx}{2}$

$latex dx =2du$

And the integral becomes:

$$\int \frac {dx}{4-x^2}=\frac{2}{4}\int \frac {du}{1-u^2}=\frac{1}{2}\int \frac {du}{1-u^2}$$

This integral also appears in tables as:

$$\int \dfrac {dx}{1-x^2}=\frac{1}{2}\ln\left |\dfrac{1+x}{1-x}\right | + C$$

$$\int \frac {dx}{4-x^2}=\frac{1}{2}\int \frac {du}{1-u^2}=\frac{1}{2}\cdot\frac{1}{2}\ln\left |\dfrac{1+u}{1-u}\right | + C$$

Finally, it is returned to the original variable:

$$\int \frac {dx}{4-x^2}=\frac{1}{4}\ln\left |\dfrac{1+\frac{x}{2}}{1-\frac{x}{2}}\right | + C=\frac{1}{4}\ln\left |\dfrac{2+x}{2-x}\right | + C$$

In general, integrals of the form $latex \int \frac {dx}{a^2-x^2}$, are solved as follows:

$$\int \frac {dx}{a^2-x^2}=\frac{1}{2a}\ln\left |\dfrac{a+x}{a-x}\right | + C$$

Alternatively, by properties of logarithms:

$$\int \frac {dx}{x^2-a^2}=\frac{1}{2a}\ln\left |\dfrac{x-2}{x+a}\right | + C$$

At first glance, this integral doesn’t look much like the previous ones, because of the $latex 6x$ term, however, by completing the square in the denominator, an integral like those in examples 8 or 9 can be obtained:

$$\int \frac {dx}{x^2+6x+5}=\int \frac {dx}{(x^2+2\cdot 3x+9)-9+5}$$

Note that with this algebraic manipulation, the denominator is not altered, but can be rewritten in a completely equivalent way, as follows:

$$\int \frac {dx}{x^2+6x+5}=\int \frac {dx}{(x+3)^2-4}$$

Then, consider:

$latex u=x+3$

$latex du=dx$

$latex a=2$

The proposed integral results as the last variant of the previous example, in which case:

$$\int \frac {dx}{x^2+6x+5}=\int \frac {dx}{(x+3)^2-4}$$

$$=\int \frac {du}{u^2-4}=\frac{1}{2}\ln\left |\dfrac{u-2}{u+2}\right | + C$$

And it is very simple to substitute back and get:

$$\int \frac {dx}{x^2+6x+5}=\frac{1}{2}\ln\left |\dfrac{(x+3)-2}{(x+3)+2}\right | + C$$

$$\int \frac {dx}{x^2+6x+5}=\frac{1}{2}\ln\left |\dfrac{x+1}{x+5}\right | + C$$