# Evaluate the indefinite integral. 1. \int \frac{5z^{2}}{(z^{3}+2)^{3}}dz\\ 2. \int...

## Question:

Evaluate the indefinite integral.

{eq}1. \int \frac{5z^{2}}{(z^{3}+2)^{3}}dz\\ 2. \int \frac{x}{(x^{2}+1)^{2}}dx\\ 3. \int e^{x}\sqrt{1+e^{x}}dx\\ 4. \int \frac{e^{v}}{(2e^{v}+5)^{2}}dv\\ 5. \int \frac{(\ln x)^{2}}{x}dx\\ 6. \int \frac{4}{x(\ln x)^{2}}dx\\ 7. \int te^{2t^{2}}dt\\ 8. \int p^{2}e^{p^{3}+2}dp {/eq}

## Indefinite Integrals By u-Substitution method.

We use the substitution method to find the antiderivatives of the integrals.

The substitution method is also called changing variables. Sometimes some problems start with a complicated function,

then we need to change its variable to make it a success.

The formula is:

{eq}\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ {/eq}

Here we have to evaluate the given indefinite integrals:

Part 1.)

{eq}\displaystyle\int \dfrac{5z^2}{(z^3+2)^3}dz\\\\ {/eq}

Applying u-substitution method we get:

{eq}z^3+2=t\\\\ 3z^2\ dz=dt\\\\ z^2\ dz=\dfrac{dt}{3}\\\\ {/eq}

Therefore integrating:

{eq}=5\displaystyle\int \dfrac{1}{t^3}\cdot \dfrac{dt}{3}\\\\ =\dfrac{5}{3}\displaystyle\int t^{-3}dt\\\\ =\dfrac{5}{3}\left [ \dfrac{-1}{2t^2} \right ]+c\\\\ =\dfrac{-5}{3}\left ( \dfrac{1}{2(z^3+2)^2} \right )+c\\\\ {/eq}

Here c denotes the integration constant.

Part 2.)

{eq}\displaystyle\int \dfrac{x}{(x^2+1)^2}dx\\\\ {/eq}

Applying u-substitution method we get:

{eq}x^2+1=z\\\\ 2x\ dx=dz\\\\ x\ dx=\dfrac{dz}{2}\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int \dfrac{1}{z^2}\cdot \dfrac{dz}{2}\\\\ =\dfrac{1}{2}\left ( \dfrac{-1}{z} \right )+c\\\\ =\dfrac{-1}{2(x^2+1)}+c\\\\ {/eq}

Here c denotes the integration constant.

Part 3.)

{eq}\displaystyle\int e^x\sqrt {1+e^x}dx\\\\ {/eq}

Applying u-substitution method we get:

{eq}1+e^x=z\\\\ e^x\ dx=dz\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int \sqrt z\cdot dz\\\\ =\dfrac{z^{\frac{3}{2}}}{\frac{3}{2}}+c\\\\ =\dfrac{2}{3}(1+e^x)^{\frac{3}{2}}+c\\\\ {/eq}

Here c denotes the integration constant.

Part 4.)

{eq}\displaystyle\int \dfrac{e^v}{(2e^v+5)^2}dv\\\\ {/eq}

Applying u-substitution method we get:

{eq}2e^v+5=z\\\\ 2e^v\ dv=dz\\\\ e^v\ dv=\dfrac{dz}{2}\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int \dfrac{1}{z^2}\cdot \dfrac{dz}{2}\\\\ =\dfrac{1}{2}\left ( \dfrac{-1}{z} \right )+c\\\\ =\dfrac{-1}{2(2e^v+5)}+c\\\\ {/eq}

Here c denotes the integration constant.

Part 5.)

{eq}\displaystyle\int \dfrac{(lnx)^2}{x}dx\\\\ {/eq}

Applying u-substitution method we get:

{eq}lnx=z\\\\ \dfrac{1}{x}dx=dz\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int z^2\ dz\\\\ =\dfrac{z^3}{3}+c\\\\ =\dfrac{1}{3}(lnx)^3+c\\\\ {/eq}

Here c denotes the integration constant.

Part 6.)

{eq}\displaystyle\int \dfrac{4}{x(lnx)^2}dx\\\\ {/eq}

Applying u-substitution method we get:

{eq}lnx=z\\\\ \dfrac{1}{x}dx=dz\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int \dfrac{4}{z^2}dz\\\\ =4\left ( \dfrac{-1}{z} \right )+c\\\\ =\dfrac{-4}{lnx}+c\\\\ {/eq}

Here c denotes the integration constant.

Part 7.)

{eq}\displaystyle\int te^{2t^2}dt\\\\ {/eq}

Applying u-substitution method we get:

{eq}2t^2=z\\\\ 2\cdot 2t\ dt=dz\\\\ t\ dt=\dfrac{dz}{t}\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int e^z\cdot \dfrac{dz}{4}\\\\ =\dfrac{1}{4}\left [ e^z \right ]+c\\\\ =\dfrac{e^{2t^2}}{4}+c\\\\ {/eq}

Here c denotes the integration constant.

Part 8.)

{eq}\displaystyle\int p^2{e^{p^3+2}}dp\\\\ {/eq}

Applying u-substitution method we get:

{eq}p^3+2=z\\\\ 3p^2\ dp=dz\\\\ p^2\ dp=\dfrac{dz}{3}\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int e^z\cdot \dfrac{dz}{3}\\\\ =\dfrac{1}{3}\left [ e^z \right ]+c\\\\ =\dfrac{1}{3}\left [ e^{p^3+2} \right ]+c\\\ {/eq}

Here c denotes the integration constant.