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

Power rule of integration: ∫x^{n} dx = $\dfrac{x^{n+1}}{n+1}$+CIntegrations of constants: ∫k dx = kx+C ∫sin x dx = -cosx+C ∫cos x dx = sin x+C ∫sec ^{2} x dx = tan x+C∫cosec ^{2} x dx = -cot x+C∫sec x tan x dx = sec x+C ∫cosec x cot x dx = -cosec x+C ∫e ^{x} dx = e^{x}+C∫a ^{x} dx = $\dfrac{a^x}{\ln a}$+C; a(≠1)>0.∫$\dfrac{dx}{x}$ = ln|x| +C |

**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

Evaluate ∫6xQuestion 1:^{2} dx |

*Answer:*

By the power rule of integration,

∫6x^{2} dx = 6 x^{2+1}/(2+1) + C

= 6 x^{3}/3 + C

= 2x^{3} + C, where C is an integration constant.

Question 2: Evaluate ∫(x^{2}+x+1) dx |

*Answer:*

∫(x^{2}+x+1) dx = ∫x^{2} dx + ∫x dx + ∫dx

= x^{3}/3 + x^{2}/2 + x + C

= 2x^{3} + C, where C is an integration constant.

Question 2: Evaluate ∫1/x^{2} dx |

*Answer:*

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

∫1/x^{2} 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

Evaluate ∫sin 2x dxQuestion 4: |

*Answer:*

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.

**Also Read:**

Evaluate ∫cosQuestion 5:^{2}x dx |

*Answer:*

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

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

∴ ∫cos^{2}x 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

Evaluate ∫eQuestion 6:^{2x} dx |

*Answer:*

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

∴ ∫e^{2x} 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.

Evaluate ∫(eQuestion 7:^{x}+e^{-x}) dx |

*Answer:*

∫(e^{x}+e^{-x}) dx

= ∫e^{x} dx + ∫e^{-x} dx

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

Evaluate ∫2Question 8:^{x} dx |

*Answer:*

We know that ∫a^{x} 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

∫2^{x} dx = $\dfrac{2^x}{\ln 2}$+C.

## Logarithmic Integration Examples

Evaluate ∫logx dxQuestion 9: |

*Answer:*

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.

Evaluate ∫$\dfrac{\log x}{x}$ dxQuestion 10: |

*Answer:*

Put z=logx

So dz = dx/x.

∴ ∫$\dfrac{\log x}{x}$ dx

= ∫z dz

= z^{2}/2 +C

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