Prove that: ∫ (1)/(a²-x²)dx = (1)/(2a)log|(a+x)/(a-x)| + c.

Question 229 of 255

\displaystyle\int \dfrac{1}{a^2-x^2}\,dx = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right| + c

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 4mImportance★★★★★

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Decompose $\dfrac{1}{a^2-x^2}$ into partial fractions using $a^2-x^2=(a-x)(a+x)$ , then integrate term by term.

Since $a^2-x^2 = (a-x)(a+x)$ , decompose into partial fractions:

$\frac{1}{a^2-x^2} = \frac{1}{(a-x)(a+x)} = \frac{1}{2a}\left[\frac{1}{a-x}+\frac{1}{a+x}\right]$

(Check: $\dfrac{1}{2a}\left[\dfrac{1}{a-x}+\dfrac{1}{a+x}\right] = \dfrac{1}{2a}\cdot\dfrac{(a+x)+(a-x)}{(a-x)(a+x)} = \dfrac{1}{2a}\cdot\dfrac{2a}{a^2-x^2}=\dfrac{1}{a^2-x^2}$ ✓)

Integrating both sides: …

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Q.Prove that: $\displaystyle\int \dfrac{1}{a^2-x^2}\,dx = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right| + c$ .

Concept understanding — Integrals of the Form ∫ dx/(ax²+bx+c) and ∫ dx/√(ax²+bx+c)

Q.Prove that: ∫1a2−x2 dx=12alog⁡∣a+xa−x∣+c\displaystyle\int \dfrac{1}{a^2-x^2}\,dx = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right| + c∫a2−x21​dx=2a1​log​a−xa+x​​+c.

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Q.Prove that: $\displaystyle\int \dfrac{1}{a^2-x^2}\,dx = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right| + c$ .

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