Integration is known as the inverse process of derivatives, also called anti-derivative. There are many techniques to find integrations. In this article, we will provide a few examples of integrations with detailed solutions.

## Integration Formulas

Integration: Definition, Formulas, Properties, Examples

## Integration Examples and Solutions

We will solve a few integrals here depending upon the type of the integrands. They will be categorized as follows:

## Power Rule of Integration Examples

By the power rule of integration,

∫6x2 dx = 6 x2+1/(2+1) + C

= 6 x3/3 + C

= 2x3 + C, where C is an integration constant.

∫(x2+x+1) dx = ∫x2 dx + ∫x dx + ∫dx

= x3/3 + x2/2 + x + C

= 2x3 + C, where C is an integration constant.

Note that 1/x2 can be written as x-2 by the rule of indices. We have:

∫1/x2 dx = ∫x-2 dx

= x-2+1/(-2+1) + C

= x-1/-1 + C

= -1/x + C, where C is an integration constant.

## Trigonometric Integration Examples

We will use the substitution method to integrate. Let us put

z=2x ⇒ dz=2 dx ⇒dx=dz/2.

Thus, ∫sin 2x dx = ∫sin z dz/2

= (1/2)∫sin z dz = (1/2) (-cos z) + C

= $-\dfrac{\cos 2x}{2}$ + C as z=2x.

Integration of mod x

Integration of Root x

We will use the following trigonometric formula: cos 2x = 2cos2x -1. Thus, we have

cos2x = $\dfrac{1+\cos 2x}{2}$

∴ ∫cos2x dx = $\dfrac{1}{2} \int (1+\cos 2x) dx$

= $\dfrac{1}{2} (x+\dfrac{\sin 2x}{2})$ as we know that ∫cos mx dx = $\dfrac{\sin mx}{m}$ +C

= $\dfrac{x}{2} +\dfrac{\sin 2x}{4}$

## Exponential Integration Examples

Put 2x=z. so we have dx=dz/2.

∴ ∫e2x dx = $\int e^z \dfrac{dz}{2}$

= $\dfrac{1}{2} \int e^z dz$

= $\dfrac{1}{2} e^z$ +C

= $\dfrac{1}{2} e^{2x}$ +C as z=2x.

∫(ex+e-x) dx

= ∫ex dx + ∫e-x dx

= ex – e-x +C as we know that ∫emx dx = $\dfrac{e^{mx}}{m}$ +C.

We know that ∫ax dx = $\dfrac{a^x}{\ln a}$ +C; a(≠1)>0. Here ln denotes the natural logarithm, that is, the logarithm with base e.

So applying the above formula for a=2, we obtain that

∫2x dx = $\dfrac{2^x}{\ln 2}$+C.

## Logarithmic Integration Examples

To find the integration of logx, we will use the integration by parts formula:

∫uv dx = u∫v dx $-\int [\frac{du}{dx}( \int v \, dx )]\, dx$, where u, v are functions of x.

Put u=logx, v=1

∴ ∫logx dx

= ∫(logx × 1) dx

= log x ∫dx – ∫x [d/dx(logx)] dx +C

= xlog x – ∫x (1/x) dx +C

= xlog x – ∫dx +C

= xlog x – x +C

So the integration of logx is equal to x(logx-1)+C.

Put z=logx

So dz = dx/x.

∴ ∫$\dfrac{\log x}{x}$ dx

= ∫z dz

= z2/2 +C

= (log x)2/2 +C, where C is an integral constant.

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