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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 51 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 51 Maths Textbook Solution.

Answers (1)

Answer:

I=\frac{1}{2} \ln \left(\mathrm{x}^{2}+4 \mathrm{x}+5\right)-\tan ^{-1}(\mathrm{x}+2)+\mathrm{c}

Given:

\int \frac{x+1}{x^{2}+4 x+5} d x

Hint:

To solve this equation we will make term into differential form.

Solution: 

\int \frac{x+1}{x^{2}+4 x+5} d x

  I=\int \frac{x+2-1}{x^{2}+4 x+5} d x                        

I=\int \frac{x+2}{x^{2}+4 x+5} d x-\int \frac{d x}{x^{2}+4 x+5}

\text { Let } x^{2}+4 x+5=t

(2 x+4) d x=d t

(x+2) d x=\frac{d t}{2}

I=\int \frac{\frac{d t}{2}}{t}-\int \frac{d x}{x^{2}+4 x+4+1}

I=\frac{1}{2} \int \frac{d t}{t}-\int \frac{d x}{(x+2)^{2}+1}

I=\frac{1}{2} \ln (t)-\int \frac{d u}{u^{2}+1} \quad[\because x+2=u, d x=d u]

I=\frac{1}{2} \ln \left(x^{2}+4 x+5\right)-\tan ^{-1} u+c

I=\frac{1}{2} \ln \left(x^{2}+4 x+5\right)-\tan ^{-1}(x+2)+c

 

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